Math 209 (Fall 2007) Calculus III. Solution #5. 1. Find the minimum and maximum values of the following functions f under the given constraints:

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1 Math 9 (Fall 7) Calculus III Solution #5. Find the minimum and maximum values of the following functions f under the given constraints: (a) f(x, y) 4x + 6y, x + y ; (b) f(x, y) x y, x + y 6. Solution: (a) Set g(x, y) x + y, so that f(x, y) 4, 6 and g(x, y) x, y. Find those x, y, λ R that satisfy 4 λx, 6 λy, and x + y. If λ, the both x and y by the first two equations, which contradicts the third one. Hence, λ most hold, so that x λ and y λ. Plugging this into the third equation yields x + y 4 λ + 9 λ λ, so that λ ± and thus x ± and y ±. Hence, (, ) and (, ) are the points to be tested. Since f(, ) 6 and f(, ) 6, the minimum value is 6 and the maximum is 6. (b) Set g(x, y) x + y, so that f(x, y) xy, x and g(x, y) x, 4y. Find those x, y, λ R such that xy λx, x λ4y, and x + y 6. Suppose that x. Then λ or y by the second equation; by the third equation, y cannot occur in this case (if x ), so that λ. From the third equation, we conclude that y ±. Suppose that x. Then λ y by the first equation and thus x 4y by the second one. Plugging into the third equation yields 6y 6 and thus y ±. By x 4y, this means that for either value of y, we have x ±.

2 All in all, we have to test the following points: (, ), (, ), (, ), (, ), and (, ). Since f (, ), f(, ) f(, ) 4 and f(, ) f(, ) 4, the minimum value is 4 and the maximum value is 4.. Find the minimum and the maximum value of f(x, y) x + y 4x 5 on the closed disc x + y 6. Solution: We first determine the minimum and the maximum value of f(x, y) on the boundary of the disc, i.e., under the constraint g(x, y) : x + y 6. We have f(x, y) 4x 4, 6y and g(x, y) x, y. etermine those x, y, λ R such that 4x 4 λx, 6y λy, and x + y 6. Suppose that y. Then the third equation yeilds x ±4. Suppose that y. The second equation then yields λ ; plugging into the first one, we obtain x 4 6x and thus x. From the third equation, we conclude that y ±. ( ) We thus have to test f at the following points (4, ), ( 4, ),,, and ( ) ( ),, and obtain: f(4, ), f( 4, ) 4, f, ± 47. To check for possible candidates for a minimum or maximum value in the interior of the disc, i.e., with x + y <, determine the stationary points of f there. Since 4x 4 6y x and y. There is only one stationary point, namely (, ). Since f(, ), we conclude that 47 is the maximum and the minimum value of f on the closed disc.. Find the mimimum and maximum value of f(x, y, z) xyz under the constraint that x + y + z. Solution: Set g(x, y, z) : x + y + z, so that f(x, y, z) yz, xz, xy and g(x, y, z) x, y, z. etermine those x, y, z, λ R such that yz λx, xz λy, xy λz, and x + y + z. Since f(x, y, z) if any of x, y, or z is equal to zero, we shall suppose that all of x, y, and z are non-zero.

3 We thus obtain that λ yz x xz y xy z. Multiplying the first of these equalities with 4xy, then yields y z x z and thus y x ; similarly, we obtain that z x. Plugging into the constraint yields x and thus x ±. Hence, we have to test f at the following points: (,, ), (,, ), (,, ), (,, ), (,, ), (,, ), (,, ), and (,, ). At each of those points, the value of f is either if there is an even number of negative coordinates or if there is an odd number of negative coordinates. The maximum value is thus and the minimum value is. 4. The plane x + y + z intersects the paraboloid z x + y in an ellipse. Find the points on this ellipse that are closed and farthest, respectively, from the origin. Solution: The distance of any point (x, y, z) from the origin is x + y + z. We thus have to find the minimum and the maximum of f(x, y) : x + y + z under the constraints g(x, y, z) and h(x, y, z), where We have g(x, y, z) : x + y + z and h(x, y, z) : x + y z. f(x, y, z) x, y, z, g(x, y, z),,, and h(x, y, z) x, y,. etermine x, y, z, λ, µ R such that the following equations are satisfied: x λ + µx, () y λ + µy, () z λ µ, () x + y + z, (4) x + y z. (5) Subtract () from (), and obtain that (x y) µ(x y). Suppose that x y; division by (x y) then yields µ. From (), we conclude that λ, so that by () z, which is impossible by (5). Consequently, x y must hold. Then (4) and (5) become x + z and x z,

4 so that x x. Since x + x (x )(x + ), we have x + x if and only if x or x. The two points to check are thus (,, ) and (,, ). Clearly, the second point has a larger distance from the origin, namely 6, then first one ( ). The first point thus is the one with the minimum, the second one the one with the maximum distance. 5. Use polar coordinates to evaluate the double integral G f(x, y) da for the following f(x, y) and : (a) f(x, y) x + y, the region to the left of the y-axis between the circles x + y and x + y 4; (b) f(x, y) 4 x y, {(x, y) R : x + y 4, x }; (c) f(x, y) ye x, the region in the first quadrant enlosed by the circle Solution: x + y 5. (a) In rectangular coordinates, we have which becomes {(x, y) R : x, x + y 4}, { [ π (θ, r) : θ, π in polar coordinates. We thus obtain: x + y da π π ( π ] }, r [, ] (r cos θ + r sin θ)r dr dθ r (cos θ + sin θ) dr dθ cos θ + sin θ dθ [sin θ cos θ] θ π θ π ( + ) 4 [ r ) ( ] r ( 8 r ) ) r dr 4

5 (b) Since we obtain: { [ (θ, r) : θ π, π ] }, r [, ], 4 x y da π π ( π 4 r r dr dθ dθ ) ( r 4 r dr r ) 4 r dr π u du, u 4 r, 4 π [ ] u4 u 8π. u (c) In polar coordinates, we have It follows that ye x da { (θ, r) : θ r sin θ e r cos θ r dr dθ [, π ] }, r [, 5]. r sin θ e r cos θ dθ dr. To evaluate the inner integral, we substitute u r cos θ, so that du r sin θ dθ and thus This yields: r sin θ e r cos θ dθ ye x da r r re u du r e u du r(e r ). r(e r ) dr re r dr r dr [re r ] r5 r e r dr 5e 5 e e 5. r dr 5

6 6. Use polar coordinates to find the volume of the solid inside the sphere x +y +z 6 and outside the cylinder x + y 4. Solution: The sphere x +y +z 6 intersects the xy-plane in the circle x +y 6, so that the volume in question is computed as V 6 x y da, where or, in polar coordinates, We hence obtain: {(x, y) R : 4 x + y 6} {(r, θ) : θ [, π], r [, 4]}. V 4π π π 4 4 [ 4π u 4π π. 6 r r dr dθ r 6 r dr u du, u 6 r, u du ] u u 6