Math 326 Assignment 3. Due Wednesday, October 17, 2012.

Transcription

1 Math 36 Assignment 3. Due Wednesday, October 7, 0. Recall that if G(x, y, z) is a function with continuous partial derivatives, and if the partial derivatives of G are not all zero at some point (x 0,y 0,z 0 ), then the gradient vector is perpendicular to the level surface G(x,y,x) = G(x 0,y 0,z 0 ) at the point (x 0,y 0,z 0 ). Find the points (x,y,z) on the paraboloid z = x +y at which the normal line to the surface is the line from the origin to the point (x,y,z). (At such a point on the surface, the vector xi+yj+zk is a scalar multiple of the gradient of the function G(x,y,z) = z x y at (x,y,z), i.e., xi+yj +zk = λ (x,y,z) G(x,y,z), for some scalar λ. The points (x,y,z) on the paraboloid have z-coordinate z = x +y ). Solution: Upon computing the gradient of F, the equation xi+yj+zk = λ (x,y,z) F(x,y,z) is the equation: xi+yj +zk = xλi yλj +λk, which gives the three component equations: (i). x = xλ (j). y = yλ (k). z = λ, together with the level surface (or graph) equation z = x +y. (a). If x and y equal 0, then the points (0,0,z), (with λ = z), satisfy equations (i),(j),(k). Thus, all points on the z-axis satisfy the gradient equation. (b). If x or y is nonzero, then by equation (i) or (j), λ =, and all points in the horizontal plane (x,y, ) satisfy the gradient equation. When we combine the points (0,0,z) in (a) with the level surface equation z = x +y, we get a point (0,0, ) on the paraboloid where the line from the origin to the point is perpendicular to the surface, namely, the point at the base of the paraboloid. When we combine the points (x,y, ) in (b) with the level surface equation z = x +y, we get the points (x,y, ) where x + y =, a horizontal circle of radius at height z =. The line from the origin to each of those points is perpendicular to the paraboloid.

2 Lagrange s method for finding extreme values. Suppose that F(x, y.z) is a function of three independent variables, whose first partial derivatives are continuous functions. Suppose that you are looking at a level surface S with equation G(x,y,z) = c for some function G(x,y,z), with continuous first partial derivatives. Suppose that the level surface S is bounded, i.e., it is contained in some sphere in 3-space. Considering just the values of F at the points of the surface S, there will be a largest value and a smallest value. That conclusion need not hold when S is an unbounded surface. Lagrange gave a method that will determine the points on S where the biggest and smallest values occur (when there are such values!). At a point P of S where the biggest (or smallest) value occurs, either the gradient vector p G of G is the 0-vector at the point, or P G is nonzero and P F is a scalar multiple of P G. (We will show that fact in class.) Lagrange s method for finding the points of the level surface G(x, y, z) where F has its biggest and smallest values (among its values on the surface) is the following procedure: (i). Find the points P on the level surface G(x,y,z) = c at which P G = 0 if any. (ii). Find the points P on the level surface G(x,y,z) = c at which P G is a nonzero vector and where the vector P F is a scalar multiple of the vector P G, i.e., where the equation P F = λ P P G has a scalar solution λ P. (iii). Compute the values of F at the points in. and. The largest and smallest among those values are the largest and smallest values of F on the surface G(x,y,z) = c. You will be searching for points (x,y,z) that satisfy the equation G(x,y,z) = c and the equation (x,y,z) F = λ (x,y,z) G, for some value of λ. Since the vector gradient equation breaks up into its 3 component equations, we want points (x, y, z) that satisfy the four equations: (i). F ((x,y,z)) = λ G ((x,y,z)). x x (ii). F y ((x,y,z)) = λ G y ((x,y,z)). F (iii). ((x,y,z)) = λ G ((x,y,z)). z z (iv). G(x,y,z) = c. The three equations (i)-(iii) depend on four variables x,y,z,λ. You should replace those three equations with two equations (i) and (ii) that depend just on the three variables x,y,z, giving you three equations (i),(ii),(iv) in three variables x,y,z to solve.

3 . A rectangular box lies in the first octant, with one corner at the origin and the diagonally opposite corner at (x,y,z), a point on the plane x + y + z =, with a,b,c all positive, and a b c x,y,z all positive. Find the maximum volume among such boxes. Solution. The volume of the box with one corner at (0,0,0) and the other corner at (x,y,z) is F(x,y,z) = xyz. We want to maximize F(x,y,z) = xyz on the plane x + y + z =. In a b c Lagrange s equation F = λ G, G(x,y,z) = x + y + z. a b c Lagrange s equation is (yz,xz,xy) = λ( a, b, c ), and the three component equations are yz = λ a, xz = λ b, xy = λ c. Hence, λ = ayz = bxz = cxy. Solving for y in terms of x via the equation ayz = bxz, by canceling the positive value z, we see that y = b x. Solving for z in terms of x via the equation ayz = cxy, by canceling a the positive value y, we see that z = c x. Thus, Lagrange s equation determines the points a (x, bx, cx ). Substituting those points in the equation of the plane (the constraint), we get a a x + x + x =. Hence, x = a, and using the equations for y and z as functions of x, we see a a a 3 that y = b and z = c. Thus, Lagrange s formula gives the single point 3 3 (a, b, c ). The value of F(x,y,z) = xyz at that point is abc, which is the maximum volume of the boxes Apply Lagrange s method to find the extreme values of x + y + z subject to the constraint x + y + z =, when a > b > c > 0. a b c Solution. Let F(x,y,z) = x +y +z, and G(x,y,z) = x is (x,y,z) = λ( x y z a, b, c ). Hence, we get the three equations x = x y z aλ,y = bλ,z = c λ, which we can write as + y + z a b x(a λ) = 0,y(b λ) = 0,z(c λ) = 0. c. The Lagrangeequation Hence, if x 0, then λ = a, and if y 0, then λ = b, and if z 0, then λ = c. But since a, b, c are all distint, those values for λ are in conflict unless two of x,y,z equal 0. If x is nonzero and y and z are zero, substituting (x,0,0) in the constraint equation (the level surface equation), we see that x = a. The value of F(x,y,z) = x +y +z at (x,0,0), when x = a, is a. Arguing in the same way when the y-coordinate is nonzero, we get the value b, and when the z-coordinate is nonzero, we get the value c. Since a > b > c > 0, we see that a is the largest value of F on the ellipsoid, and c is the smallest value.

4 4. Find the minimum distance from (0,0,c) to the cone z = x a > b > 0. a + y b, with c > 0 and Solution. Let F(x,y,z) = x + y + (z c), which measures the distance-squared from (0,0,c)to(x,y,z)andlet G(x,y,z) = z x y = 0betheconstraint equation. Lagrange s a b equation says that (x,y,(z c)) = λ( x a, y b,z). Hence, x = xλ, y = yλ a b and (z c) = λz. Let us write those equations as x(a +λ) = 0,y(b +λ) = 0,( λ)z = c. x and y cannot both be nonzero. If they were, then the first two equation would give the conflicting values λ = a = b for λ. Hence, at least one of x and y must be 0. If both x and y are zero, then by the constraint, the z-coordinate is also 0, which gives the origin (0,0,0). But (0,0,0) does not satisfy the equation ( λ)z = c above since c > 0. So, (0,0,0) is not a point we need to consider. Next, suppose that y is not zero and x is 0. Then we can cancel y in the equation y = yλ and conclude that λ = b. Then by the equation ( λ)z = c, we conclude that b z = c. In the same way, if x is nonzero and y = 0, we conclude that z = c. Thus, the +b +a c c points that satisfy Lagrange s equation are (0, y, ) and (x,0, ), with x and y nonzero. +b +a c Substituting the points (0, y, ) in the constraint, we find that ( c ) 0 y = 0. Hence, +b +b b y = ( bc ) c. Hence, the value of F(x,y,z) at (0,y, ) (with y = ( bc ) ) equals +b +b +b 0 bc c +( +( +b ) +b c) = b c +b. bc Hence, the distance from (0,0,c) to those points equals c +b. From the points (x,0, ac we get, in the same way, the distance ac +a. Now we have to decide which of bc +b is lesser. That follows from the fact that bc +b is lesser when a > b > 0. increasing function of x for x > 0, since the function has a positive derivative. ), +a +a and cx +x is an

5 5. Find the points among which occur the points of maximum and minimum value of F(x,y,z) = x+y 3 +z 3 on the sphere G(x,y,z) = x +y +z =, using Lagrange s method. If you are able to, using a computer or calculator, compute the values of F at the points that you find to see which value is largest and which is smallest. (There are 4 points to be found via Lagrange s method.) Solution: We have: F = i+3y j +3z k and G = xi+yj +zk. G equals the zero vector only at (0,0,0), which is not a point of the level surface G =. Hence, at no point of the level surface is the gradient of G equal to 0. We need to find the points P = (x 0,y 0,z 0 ) satisfying Lagrange s equation P F = λ P G, for some scalar λ. The i,j,k component equations of that vector equation are = xλ, 3y = yλ, 3z = zλ. The first equation tells us that x 0 and that λ =. Substituting the value of λ in second x and third equations, we get the equations 3y = y x and 3z = z x. Case. y and z are nonzero. In that case, we can cancel y and z from the second and third equations to get the equations 3y = and 3z =, i.e., y = = z. Thus, we find the points x x 3x (x,, ), where x is nonzero, which satisfy Lagrange s equation. 3x 3x We need to find out which of those points are on the level surface. Substituting the points ((x,, ) in the level surface equation 3x 3x x +y +z =, we find that x + =. Therefore, 9x 9x 4 9x + = 0. The roots of that equation are x = 9± = 9±3 8. Hence, x = or 3 x = 3 four points: (x,y,z) = (x, 3x, 3x ) = ±(,, );±(, , and so, x = ± 3 or x = ± 3. Those four values of x determine 6, 6 ).

6 Case. Suppose that z = 0 and y is nonzero. In that case, we are looking at a point (x,y,0) that satisfies the equation y =, as in case. Thus, we have the points (x,,0). 3x 3x Substituting those points in the level surface equation x + y + z =, we find that x + =. Hence, 9x 4 9x + = 0, and so, 9x x = 9± 45 8 = 3± ± Therefore, x = ± 5. Those four values of x give four points of the form (x,,0). 6 3x Case 3. Suppose that y = 0 and z is nonzero. Just as in Case, we find the points (x,0, 3± and the values x = ± 5. Those four values of x give four points of the form (x,0, ). 6 3x Case 4. Suppose that y and z are both zero. Then we are looking at points (x,0,0) that satisfies the equation G(x,y,z) = x +y +z =, i.e., x =. Hence, x = ± and we get two points (±,0,0). Altogether, we have found 4 points, among which occur all the points at which F has its maximum value on the sphere, as well as all the points at which F has its minimum value on the sphere. We will not compute the values of F at the 4 points in order to find the maximum value, since it is not easy to compare the values of F at the 4 points to see which is largest, without the aid of a calculator. F(x,y,z) does have maximum and minimum values on a sphere since a continuous function has maximum and minimum values on a closed and bounded surface. 3x ),