Multivariate Calculus Review Problems for Examination Two

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1 Multivariate Calculus Review Problems for Examination Two Note: Exam Two is on Thursday, February 28, class time. The coverage is multivariate differential calculus and double integration: sections 13.3, 13.4, 13.5, 13.6, 13.7, 13.8 and 14.1, Group One (1) Suppose that the portion of a tree that is usable for lumber is a right circular cylinder. If the usable height of a tree increases 3 feet per year and the usable radius increases 1 foot per year, how fast is the volume of usable lumber increasing when the usable height of the tree is 25 feet and the usable radius is 2 feet? Comments: Let V πr 2 h. Then by the Chain Rule we find that dv dt V dr r dt + V dh h dt, or dv dt V r + 3 V dv. Hence the derivative h dt equals 2πrh + 3 πr2 1π + 12π 112π. (2) (a) Find a normal vector to the surface given by the ellipsoid 3x 2 +2y 2 +z 2 6 at the point (1, 1, 1). (b) Find parametric equations for the tangent line to the curve of intersection of the two surfaces given by 3x 2 + 2y 2 + z 2 6 and 3xy + 2yz 1 at the point (1, 1, 1). Comments: (a) Let f(x, y, z) 3x 2 + 2y 2 + z 2. The desired normal vector is n f(1, 1, 1) but f 6x, 4y, 2z so n 3, 2, 1. (b) A parallel vector t to the tangent line is given by the cross product of the gradients of the functions f from part (a) and g(x, y, z) 3xy + 2yz. Let n 2 g(1, 1, 1) so g 3y, 3x + 2z, 2y so n 2 3, 1, 2. Then t n 1 n 2 5, 9, 3. The parametric equations for the tangent line are x 1 + 5t, y 1 5t, z 1 3t. (3) Find a point on the surface z 16 4x 2 y 2 at which the tangent plane is perpendicular to the line x 3 + 4t, y 2t, and z 2 t. Comments: The desired point occurs where the gradient of g(x, y, z) 16 4x 2 y 2 z is a multiple of the parallel vector v of the given line so v 4, 2, 1. But g 8, 2y, 1. We need 8x, 2y, 1 C 4, 2, 1. This occurs with C 1 and so 8x 4 or x 1/2, 2y 2 or y 1. (4) Find all saddle points of the z x 2 2y 2 6x + 8y + 3. Justify your answer. Comments: The critical point occurs when / x / y. Now / x 2x 6 so x 3 and / y 4y + 8 so y 2. The second order partials are 2 z/ x 2 2, 2 z/ y 2 4, while 2 z/ x y. Consequently det [ 2 z/ x 2 2 ] z/ x y 2 z/ x y 2 z/ y 2 8 so the critical point is a saddle point. (5) Find the volume of the solid that lies in the first octant and is bounded by the cylinder x 2 + z 2 4 and the planes z, y, and y x. Comments: The volume is given by a double integral 4 x 2 da where is the domain in the xy-plane bounded by the lines x, y, y x, and x 2. So 2 x 2 4 x 2 da 4 x 2 dydx x 4 x 2 dx u 1/2 du 8 3 1

2 (6) Evaluate the following double integral by reversing the order of integration: 2 1 y/2 Comments: 1 1/e e x2 dx dy. 2 x1 xy/2 e x2 dx dy 1 y2x y 2. Group Two e x2 dy dx 1 2xe x2 dx e x2 1 (1) The volume of a right circular cone of radius r and height h is V (1/3)πr 2 h. Show that if the height remains constant while the radius changes, then the volume satisfies V r 2V r. Comments: We first compute On the other hand, V r r 2V r 2 r ( ) 1 3 πr2 h 2 3 πrh. ( ) 1 3 πr2 h 2 3 πrh, so the equation V r 2V r holds. (2) Let T x 2 y xy 3 + 2; x r cos θ, y r sin θ. Find the partials T/ r and T/ θ. Comments: By the chain rule, we have that T r T x x r + T y y r ( 2x y 3) cos θ + ( x 2 3xy 2) sin θ ( 2r cos θ r 3 sin 3 θ ) cos θ + ( r 2 cos 2 θ 3r 3 cos θ sin 2 θ ) sin θ 2r cos 2 θ r 3 cos θ sin 3 θ + r 2 cos 2 θ sin θ 3r 3 cos θ sin 3 θ. Similarly, by the chain rule, T θ T x x θ + T y y θ ( 2x y 3) r sin θ + ( x 2 3xy 2) r cos θ ( 2r cos θ r 3 sin 3 θ ) r sin θ + ( r 2 cos 2 θ 3r 3 cos θ sin 2 θ ) r cos θ ( 2r 2 cos θ sin θ r 4 sin 4 θ ) + ( r 3 cos 3 θ 3r 4 cos 2 sin 2 θ ) (3) Let z ln(x 2 + 1), where x r cos θ. Find / r and / θ. Comments: Use the chain rule: θ dz x dx θ 2x x r cos θ ( r sin θ) r 2 cos 2 +1 ( r sin θ) 2r2 cos θ sin θ r 2 cos / r is found similarly. (4) The length, width, and height of a rectangular box are increasing at the rates of 1 in/sec, 2 in/sec, and 3 in/sec, respectively. (a) At what rate is the volume increasing when the length x is 2 in, the width y if 3 in, and the height z is 6 in? (b) At what rate is the length of the diagonal increasing at that instant? 2

3 (c) At what rate is the surface area of the box increasing at that instant? Comments: (a) The volume V xyz with V (2, 3, 6) 36. So by the chain rule dv dt V dx x dt + V dy y dt + V yz(1) + xz(2) + yz(3) (b) The diagonal of the box is given by f(x, y, z) x 2 + y 2 + z 2 so f(2, 3, 6) Note that x x x 2 + y 2 + z 2 dz dt x f(x, y, z) and, similarly, / y y/f(x, y, z) and / z/f(x, y, z) So by the chain rule: df dt dx x dt + y (1) + x y x f(x, y, z) (1) (1) (2) dy dt + (2) + dz dt (3) y f(x, y, z) (2) + (3) z f(x, y, z) (3) (c) The surface area S is given by S(x, y, z) 2(xy+xz +yz) so S(2, 3, 6) 2( ) 64. By the chain rule, ds dt S dx x dt + S dy y dt + S dz dt 2(y + z)(1) + 2(x + z)(2) + 2(x + y)(3) 2(9) + 4(8) + 6(5) 8. (5) Let z f(u) and u g(x, y). Express the partial derivative / x in terms of dz/du, u/ x, and u/ y. Comments: By the chain rule, we have x dz u du x, y dz u du y (6) Let f(x, y) x x+y (a) Find the directional derivative of the function f at P (1, ) in the direction of the point Q( 1, 1). (b) Find a unit vector u for which u f(2, 3). Comments: (a) The directional derivative of f at P in the direction of the vector a is given by a a f(p ) f(p ) a Now we find that f x, y 1 x + y x (x + y) 2, x (x + y) 2. So f(1, ), 1 and a P Q 2, 1 with a 5. So a f(p ) 1/ 5. (b) For v f(p ) to be zero, we need to find a vector v which is perpendicular to f(1, ), 1. We can simply take v i. 3

4 (7) Let z 3x 2 y 2. Find all points at which z 6. Comments: z 6x, 2y so z 36x 2 + 4y 2 6 will give an ellipse with the equation x 2 + (y/9) 2 1. (8) Show that the surfaces z f(x, y) x 2 + y 2 and z g(x, y) (1/1)(x 2 + y 2 ) + 5/2 intersect at the point (3, 4, 5) and have a common tangent plane at that point. Comments: Note that f(3, 4) and g(3, 4) 25/1+5/2 5 as well so the two surfaces do indeed intersect at the point (3, 4, 5). Let n 1 be the normal vector to the first surface and n 2 the second. Then the surfaces will share the same tangent plane provided their normal vectors are multiplies of each other. Recall that n 1 / x, / y, 1 and n 2 g/ x, g/ y, 1 Now / x x/ x 2 + y 2 and / y y/ x 2 + y 2 so at the point (3, 4) / x 3/5 and / y 4/5 and n 1 3/5, 4/5, 1. Next, g/ x x/5 and g/ y y/5 so at the point (3, 4) g/ x 3/5 and g/ y 4/5 and n 2 3/5, 4/5, 1. Since the normal vectors are identical, the tangent planes agree. (9) Find all points on the surface f(x, y, z) x 2 + y 2 z 2 1 at which the normal line is parallel to the line through the points P (1, 2, 1) and Q(4,, 1). Comments: A normal vector to the level surface f(x, y, z) x 2 + y 2 z 2 1 is given by its gradient f 2x, 2y, 2z. We need to find all points where f is proportional to the vector v P Q 3, 2, 2 ; that is, 2x, 2y, 2z c 3, 2, 2 where c is a scalar. Hence, we have the three equations: 2x 3c, 2y 2c, 2z 2c. It follows immediately that y c and z c so y z and x 3c/2. Substituting back into the level surface equation, we find that (3c/2) 2 + c 2 c 2 1 or 3c/2 ±1 or c ±2/3. Hence there are two points are the level surface (1, 2/3, 2/3) and ( 1, 2/3, 2/3). (1) Find parametric equations for the tangent line to the curve of intersection of the cone z x 2 + y 2 and the plane x + 2y + 2z 2 at the point P (1, 1, 2). [Hint: the desired tangent line is the line of intersection of two tangent planes.] Comments: A tangent vector t to the curve of intersection is given by the cross product of the n 1 that is normal to the cone with n 2 that is normal to the plane. Now n 1 / x, / y, 1 x/ x 2 + y 2, y/ x 2 + y 2, 1 and n 2 1, 2, 2. At the point P (1, 1, 2), n 1 1/ 2, 1/ 2, 1. Now n 1 n 2 2 2, 1 + 2, 3 2/2. Hence the tangent plane is given by 2 2(x 1) + (1 + 2)(y + 1) 3 2(z 2)/2. (11) Find all saddle points of the z f(x, y) x 2 + xy + y 2 3x. If there are none, verify this as well. Comments: First, find the critical points of f(x, y); that is, / x z x 2x + y 3 and / y z y x + 2y. That is, 2x + y 3 and x + 2y which has the unique solution (x, y) (2, 1). Second, we need to determine if z xx z yy z 2 xy is negative at the critical point. Now z xx 2 and z yy 2 while z xy 1. Since 3 >, there are no saddle points. In fact, (2, 1) is a relative minimum since z xx >. 3. Group Three (1) (a) Set up only the double integral that gives the volume of the solid that lies in the first octant and is bounded above by the cylinder x 2 + z 2 16, below by the plane z, and finally laterally by the two planes x 4 and y 2x. o not evaluate the integral! Comments: (a) The domain of integration in the xy-plane is a triangle bounded by the 4

5 lines x, x 4, and y 2x so its vertices are (, ), (4, ), and (4, 8). So the double 4 y2x integral is 16 x 2 dy dx. y (2) (a) Given the double integral 8 y4 yx/2 e y2 dy dx, rewrite it by reversing the order of integration. (b) Evaluate this integral. Show your work for credit. Comments: (a) The domain of integration is a triangle in the xy-plane bounded by x, y 4, and y x/2 so its vertices are (, ), (, 4), and (8, 4). Then the integral in reversed order is (b) 4 x2y x du 2y dy to get 4 x2y x e y2 dx dy 16 e y2 dx dy. 4 2y e y2 dy. Make the change of variables u y 2 with u16 e u du e u e u (3) Find parametric equations for the tangent line to the curve of intersection of the elliptic paraboloid z x 2 + 3y 2 and the plane x + y z 2 at the point P (1, 1, 4). Comments: The normal vector N 1 to the paraboloid is given by x, y, 1 2x, 6y, 1. At x 1, y 1, N 1 2, 6, 1. The normal vector N 2 to the plane is 1, 1, 1. Their cross product N 1 N 2 5, 1, 4. Hence the tangent line is x 1 + 5t, y 1 t, z 4 + 4t. (4) (a) Suppose that a function z f(x, y) is differentiable at the point P (1, 3). Suppose that f(p ) 1, x (P ) 2, and y (P ) 1. Use a linear approximation with differentials to estimate the value f(1.2, 2.97). (b) Given the function z f(x, y) 3x + y 2 and that x u 2 v 2, y 2uv, use the chain rule to find the partial derivative u. Comments: (a) The linear approximation is given by f(p ) f(p ) + x (P ) x + y (P ) y f(p ) (.3) + ( 2) (.5) (b) u x x u + y y u (5) (a) Find all the critical points of the function y (2u) + (x 2y) (2v) 2uv 2u + (u 2 v 2 4uv) 2v 6u 2 v 2v 3 8uv 2 f(x, y) xy 1 2 x2 1 3 y3. (b) Classify each critical point as a relative maximum, relative minimum, or saddle point. Comments: (a) f x y x and f y x y 2 so x y and x y 2 so y y 2. Hence y or y 1 so there are two critical points (, ) and (1, 1). (b) We need f xx 1, f yy 2y, and f xy 1. So the Hessian f xx f yy f 2 xy 2y 1. At (, ), 1 so it is a saddle point. At (1, 1), 1 and f xx 1, so it is a relative maximum. 5

6 (6) Let g(x, y, z) y 2 cos(xz), a 2i j + 2k, and P (π/2, 1, 1). (a) Find the value of the directional derivative of g at the point P in the direction of the vector a. (b) Find the maximal value of the directional derivative of g at the point P. Comments: (a) g y 2 z sin(xz), 2y cos(xz), xy 2 sin(xz) so g(π/2, 1, 1) 1,, π/2. a So a f(p ) g(p ) a 1,, π/2 2, 1, 2 /3 (2 π)/3. (b) g(p ) 1,, π/2 1 + π 2 /4. 4. Group Four (1) (a) Find the volume of the solid that lies in the first octant and is bounded above by the cylinder x 2 + z 2 1 and the planes z, y, and y x. (b) Evaluate the following double integral by reversing the order of integration: π/2 π/2 y sin(x 2 ) dx dy. Comments: (a) The domain of integration is the triangle with vertices (, ), (1, ), and (1, 1). So we have 1 x 1 1 x 2 da 1 x 2 dydx x 1 1 x 2 dx 1 2 u 1/2 du 1/3. (b) π/2 π/2 y sin(x 2 ) dx dy π/2 yx sin(x 2 ) dydx This integrates to (1 cos(π 2 /4))/2. (2) Identify all the critical points of the function π/2 f(x, y) xy x 3 y 2. x sin(x 2 ) dx 1 2 π 2 /4 sin(u) du. Classify each critical point as a relative maximum, relative minimum, or saddle point. Comments: Step 1: Find the critical points: f x y 3x 2 and f y x 2y. So we get the system of equations y 3x 2 and x 2y. This yields x 2 3x 2 or x and x 1/6. There are two critical points: (, ) and (1/6, 1/12). Step 2: classifcation: We need f xx 6x, f yy 2, and f xy 1. So f xx f yy f 2 xy 12x 1. At x, 1, so we get a saddle point. At x 1/6, >. Since f xx <, we get a relative maximum. (3) (a) Suppose that a function z f(x, y) is differentiable at the point P (2, 1). Suppose that f(p ) 3, f x (P ) 2, and f y (P ) 1. Use a linear approximation with differentials to estimate the value f(1.98, 1.1). (b) Find the point on the surface given by the graph of z x 2 + y 2 at which the tangent plane is perpendicular to the line x 1 + 3t, y 2 + 4t, z 1 + 2t. Comments: (a) f(p ) f(p ) + f x (P ) x + f y (P ) y 3 + 2(.2 + ( 1)(.1) (b) The surface normal n 2x, 2y, 1 while the parallel vector v to the line is v 3, 4, 2. We need that n cv where c is a scalar. We obtain 2x, 2y, 1 c 3, 4, 2, so c 1/2. Consequently, x 3/4 and y 1 to yield the point ( 3/4, 1, 25/16). 6

7 (4) Find parametric equations for the tangent line to the curve of intersection of the paraboloid z x 2 + y 2 and the plane 2x + y z 1 at the point P (1, 2, 5). Comments: The normal to the surface at P is n 1 2x, 2y, 1 2, 4, 1 while n 2 2, 1, 1 is normal to the plane. Their cross product provides the tangent vector i j k ,, 6. We take t 1,, 2 as the tangent vector to give the line x 1 + t, y 2, z 5 + 2t. (5) (a) Let z f(x, y) x 2 3y 2, and x r cos θ, y r sin θ. Use the chain rule to find the partial derivative / θ. (b) Consider a circular cylinder with radius r and height h. If the radius is increasing at the rate of 2 meters per second and the height is increasing by the rate of 3 meters per second, find the rate is the volume changing when the radius is 5 meters and the height is 1 meter using the chain rule. Comments: (a) The chain rule yields θ x x θ + y y θ 2x( r sin θ) 6y(r cos θ) 2r 2 cos θ sin θ 6r 2 cos θ sin θ 8r 2 cos θ sin θ. (b) Since V πr 2 h and dr/dt 2 and dh/dt 3, we find that dv dt V dr r dt + V dh h dt 2πrhdr dh + πr2 2π + 75π 95π. dt dt 5. ouble Integral Review Problems (1) Sketch the domain of integration of the following double integrals: (a) (c) 1 yx y 1 xy x (x + y 2 ) dydx, (b) (x + y 2 ) dxdy, (d) 1 y1 yx 1 x1 xy (x + y 2 ) dydx, (x + y 2 ) dxdy. (2) Evaluate the double integral xy da where is the domain bounded by the curves y x, y 6 x, and y. Comments: If we write da dydx, we need two double integrals to perform the evaluation. Note that the line y 6 x intersects y x at the point (4, 2). To see this, solve x 6 x or x (6 x) 2 which reduces to x 2 13x This quadratic factorizes as (x 4)(x 9). Since 9 lies outside the domain, the point must be (4, 2). Hence we find that 4 y x 6 y6 x xy da xy dydx + xy dydx. y 4 y If we write da dxdy, then there is only one double integral 2 x6 y xy da xy dxdy. 7 xy 2

8 (3) Find the values of the following where is the domain enclosed by the unit circle: (a) 5dA, (b) x da, (c) (3x 4y). Comments: (a) Since da Area(), we have 5dA 5π. (b,c) By symmetry, x da y da. So the answer for both parts is. (4) Find yda where is the domain in the xy-plane bounded between the line x + y 5 and the circle x 2 + y Comments: The line and circle have two intersection points at (5, ) and (, 5). So we need to compute 5 y 25 x 2 y da y dy dx 5 1 y5 x y 25 x 2 2 y2 y5 x 5 1 [ (25 x 2 ) (5 x) 2] dx 2 which is an integral of a polynomial in x. (5) Find the volume of the solid in the first octant bounded above by the paraboloid z x 2 +3y 2 and below by the plane z, and laterally by y x 2 and y x. Comments:The base of the solid in the xy-plane is bounded between the curves y x 2 and y x. We find that 1 yx (x 2 + 3y 2 ) da (x 2 + 3y 2 ) dydx yx dx (x 2 y + y 3 ) yx yx 2 dx [ 2x 3 (x 4 + x 6 ) ] dx which is easy to evaluate. (6) Find the volume of the solid that is common to the cylinders x 2 + y 2 25 and x 2 + z 2 25 that lies above the xy-plane Comments: The base of the solid is just the domain bounded by the circle x 2 + y 2 25 so the volume is given by 5 y 25 x 2 25 x 2 da 25 x 2 dy dx y 25 x 2 y 25 x 2 25 x 2 y (25 x 2 ) dx y 25 x 2 which is easy to evaluate. (7) Set up only the double integral for the volume of the solid bounded above by the paraboloid z x 2 + y 2, below by the xy-plane, and laterally by the cylinder x 2 + (y 1) 2 1. Comments: 1 1 y1+ 1 x 2 y1 1 x 2 (x 2 + y 2 ) dy dx. (8) Find the volume of the solid in the first octant bounded above by the graph of z 9 x 2, below by z, and laterally by y 2 3x. 8 dx

9 Comments: The domain in the xy-plane is bounded between the parabola y 2 3x and the line obtained from the intersection of z 9 x 2 with z so x 3. Then the volume is given by 3 x3 (9 x 2 ) da (9 x 2 ) dxdy 3 3 xy 2 /3 (9x x 3 /3) x3 xy 2 /3 dy [(27 9) (3y 2 y 6 /3 4 )] dy which is an easy integral to evaluate in y. In reverse order, we get the equivalent double integral 3 y 3x (9 x 2 ) da (9 x 2 ) dy dx y (9 x 2 )y y 3x y dx (9 x 2 )( 3x) dx 3 which again is an straightforward integral to evaluate. (9 x x 5/2 ) dx (9) Reverse the order of integration in the following double integrals: (a) (b) 4 x8 x2y Comments: (a) 2 x4 f(x, y) dxdy, (c) xy 2 f(x, y) dxdy, (b) 1 xe y x1 8 yx/2 y π π f(x, y) dxdy. f(x, y) dydx, y (c) 4 y x e y1 1 yln x y f(x, y) dydx. f(x, y) dydx, sin x (1) Consider the double integral dx dy. (a) escribe the domain in the xy-plane y x given by the limits of integration. This domain is given in terms of horizontal crosssections. (b) Write down the equivalent double integral by reversing the order of integration by giving the domain by means of vertical cross-sections. (c) Evaluate the integral in part (b). π yx sin x π Comments: The integral in reversed order is dy dx sin x dx 2. x 6. Max-Min Problems (1) Find all critical points of f(x, y) 4xy x 4 y 4 and classify them according to relative maxima, minima, and saddle points. Comments: The partial derivatives of f(x, y) are x 4y 4x3, 9 y 4x 4y3.

10 Hence y x 3 and x y 3. So y x 3. Substituting back, we get x x 9 or x(1 x 8 ). Hence x or x ±1. The corresponding y values are y, y 1, and y 1; that is, the critical points are The second order partial derivatives are 2 f x 2 12x2, The resulting 2 2 determinant is (, ), (1, 1), ( 1, 1). 2 f y 2 12y2, 2 f x y 4. 12x y 2 144x2 y At (, ), 16 so (, ) is a saddle point; at (1, 1), >. Since f xx (1, 1) 12 <, (1, 1) is a relative maximum; at ( 1, 1), is again positive and f xx ( 1, 1) is negative, so ( 1, 1) is a relative maximum. (2) Find all critical points of f(x, y) x 5 + y 5 5xy. Classify them as a relative maximum, relative minimum, or saddle point. Comments: We proceed as in the above problem. We find that f x 5x 4 5y so y x 4 and f y 5y 4 5x so x y 4. Combining, we now have x (x 4 ) 4 x 16 so x 16 x. Factoring, we obtain x(x 15 1) with solutions x and x 1 with the corresponding values of y as y and y 1. In other words, the critical points are (, ) and (1, 1). Next, we compute the second order partials: f xx 2x 3, f yy 2y 3, f xy 5. Then the 2 2 determinant is 2x y 3 4x3 y At (, ), 25 < so (, ) is a saddle point. at (1, 1), 4 25 > and f xx (1, 1) > so (1, 1) is a relative minimum. (3) Find all point (x, y, z) on the surface z 2 xy 5 that are closest to the origin. Comments: We need to minimize the distance from a point on the surface to the point (,, ); that is, x 2 + y 2 + z 2. Almost always, it is easier to minimize the distance squared: x 2 + y 2 + z 2. Since z xy on the surface, this becomes We need to find the critical points of f: 2x + y, x f(x, y) x 2 + y 2 + xy. y 2y + x. We obtain the linear system: 2x + y and x + 2y which has the only solution (, ) which is the critical point of f. Further, the second order partial derivatives of f are Hence, f xx 2, f yy 2, f xy Since f xx >, (, ) is a minimum. Since z xy 5, the corresponding z-values are ± 5. So there are two points on the surface at minimum distance (,, ± 5). 1

11 (4) Find the absolute extrema of z f(x, y) x 2 + y 2 on the domain bounded by the ellipse x 2 /4 + y 2 /9 1. Comments: This is a two-part problem. First, we need to find all the relative extrema inside the ellipse; then, second, we need to check the boundary ellipse itself for extrema. The critical points of f(x, y) x 2 + y 2 are given by f x 2x and f y 2y. So there is only one critical point (, ) with f(, ). Note that 4 and f xx 2 so it is a minimum and not a saddle point. On the boundary ellipse, x 2 cos t, y 3 sin t with t 2π. Then z (2 cos t) 2 + (3 sin t) 2 4 cos 2 t + 9 sin 2 5. We need to find its absolute extrema. Now dz dt 1 cos t sin t, so its critical points are, π/2, π, 3π/2. But z z(t) z() 4, z(π/2) 9, z(π) 4, and z(3π/2) 9. Hence, the absolute maximum of f on is 9 while the absolute minimum of f on is. (5) A rectangular box without a lid is made from 12 square meters of cardboard. Find the maximum volume of such a box. Comments: enote the length, width, and height of the box by x, y, and z, respectively. So its volume is V xyz. We can solve for z in terms of x, y since its surface area is 12; that is, 2xz + 2yz + xy 12 shows z (12 xy)/[2(x + y)]. Hence, 12 xy V V (x, y) xy 2(x + y) 12xy x2 y 2. 2(x + y) We now find its critical points where both partial derivatives are zero: ( V y2 12 x 2 2 xy ) 1/2 x (x + y) 2, V ( x2 12 x 2 2 xy ) 1/2 y (x + y) 2. These derivative are both zero if x y. This solution is not physical. So, we examine the equations: 12 x 2 2 xy 12 y 2 2 xy. Hence, x 2 y 2 or x y since x, y. Substitute this condition in 12 x 2 2 xy to obtain 12 3x 2 so x 2. Similarly, y 2. Then z (12 xy)/[2(x + y)] 1. Hence, the maximal volume V (x, y) xyz 4. (6) Find the absolute maximum and minimum values of the function f(x, y) x 2 2xy + 2y on the rectangle {(x, y) : x 3, y 2}. Comments: Since is a closed and bounded set, the function f will have absolute extrema. First we find the critical points where both partial derivatives are zero: 2x 2y, x y 2x + 2. There is only one critical point (1, 1) and the value of f there is f(1, 1) 1. Next we need to find the extrema on each of the boundary lines, say L 1, L 2, L 3, and L 4 where L 1 is on the x-axis (y ) with x 3; L 2, the vertical line x 3, with y 2;the horizontal line L 3, with y 2 and x 3; and L 4, the vertical line x and y 2. On L 1, f(x, ) x 2. This function is increasing. Its minimum value is f(, ) and maximum value is f(3, ) 9. On L 2, f(3, y) 9 4y. This function is decreasing. Its maximum value is f(3, ) 9 and mimimum value is f(3, 2) 1. 11

12 On L 3, f(x, 2) x 2 4x + 4 (x 2) 2. Its minimum value is f(2, 2) and maximum value is f(, 4) 4 since x 3. On L 4, f(, y) 2y. Its maximum value is f(, 2) 4 and minimum value is f(, ). Conclude: on the edges, the minimum value of f is while its maximum value is 9. Since the value of f at the interior critical point is 1, the absolute extrema occur on the boundary. 12