Solutions to assignment 3

Transcription

1 Math 9 Solutions to assignment Due: : Noon on Thursday, October, 5.. Find the minimum of the function f, y, z) + y + z subject to the condition + y + z 4. Solution. Let s define g, y, z) + y + z, so the problem is to find the minimum of f, y, z) subject to the constraint g, y, z) 4. We have f λ g, y, z) λ,, ); and reading this component by component we obtain λ, y λ, z λ. Plugging this into the constraint we have ) λ λ + λ + 4 λ 4 7. Thus, y 4, z, and, 4, ) is the only critical point. Now we could use the Hessian matri of f and see that it is positive definite to justify that this critical point gives the minimum. Alternatively, we can note that the function f is unbounded above even subject to the restriction) and therefore has no maimum, but it has a minimum since it is bounded below by. Therefore the minimum subject to the given restriction is f, 4, ) Find the maimum value of the function F, y, z) +y +z), subject to the constraint given by + y + z. Solution. Let s define g, y, z) +y +z, so the problem is to find the maimum of F, y, z) subject to the constraint g, y, z). We have F λ g + y + z), + y + z), + y + z)) λ, 4y, z). eading this component by component and including the restriction we get the system of equations + y + z λ + y + z λy + y + z λz + y + z. A) B) C) D)

2 Subtracting A) B) we get λ y), so either λ or y. But λ would give y z, and f,, ) is obviously not the maimum. Therefore we work with y. Subtracting B) C) we get λy z), and since we already discarded the case λ we are left with z y. Using the results in the two frames into D) we get ) y) + y + y y ± ±, z ±. It is clear that the maimum of F occurs when, y, z are all positive, or when they are all negative. Therefore the maimum value is F,, ) ) F,, ± ).. Find the maimum and minimum values of the function f, y, z) y z, subject to the constraints + y z, + z. Solution. Let s define g, y, z) +y z and h, y, z) +z, so the problem is to find the maimum of f, y, z) subject to the constraints g, y, z) and h, y, z). We have f λ g + µ h,, ) λ,, ) + µ,, 4z). eading this component by component and including the restrictions we get the system of equations λ + µ A) λ λ + 4µz + y z + z. B) C) D) E)

3 Note that B) already gives λ. Using this in A) and C) we obtain and µ z respectively. Plugging these epressions for and z into E) we get µ Now, from D) we have y z, so we get ) + µ ± µ µ). µ, z, y. µ, z, y. Since the intersection of + y z and + z is closed and bounded, all we need to do now is evaluate f at the critical points we have found. f,, ) is the maimum value, f,, ) is the minimum value. 4. Find the etreme values of the function f, y, ) y + z on the region described by the inequality + y + z. Use Lagrange multipliers to treat the boundary case. Solution. First we work in the interior: + y + z <. to find the critical points we set f. This yields y, so the only critical point in the interior is,, ). But clearly f,, ) is neither a maimum nor a minimum. It is also clear that there are no singular points. Now we work on the boundary: +y +z. Here we can define g, y, z) +y +z, so the problem is to find the etreme values of f, y, z) subject to g, y, z). We have f λ g y,, z) λ, y, z). eading this component by component and including the restriction we get y λ λy z λz + y + z. A) B) C) E) Note that C) implies z λ), so either z or λ. Case : z. Note that A) and B) imply y, and then from D) we get y. this way we get four points: ±, ±, ).

4 Case : λ. Now A) and B) imply y, and then from D) we get z ±. This way we get the two points,, ±). Since +y +z is closed and bounded, all we need to do now is evaluate the function at the points we have found: ) f,, f, ), f, ), f ),, this is the global minimum) f,, ±) this is the global maimum). 5. Use Lagrange multipliers to prove that a rectangle with maimum area, that has a given perimeter p, is a square. Solution. Let the sides of the rectangle be and y, so the area is A, y) y. The problem is to maimize the function A, y) subject to the constraint g, y) + y p p > is a fied number). We have A λ g y, ) λ, ). eading this component by component we get { y λ λ y so the rectangle with maimum area is a square with side length p 4.. Evaluate y + dy. Solution. Since we are integrating with respect to y, the letter in the integrand is treated as a constant. We have y + dy dy y arctany) y y + arctan) arctan)) arctan).

5 7. Calculate the iterated integral + y) ddy. Solution. We have + y) ddy + y) ) dy [ + y) + y ] dy ln + y) y y + lny) y y ln) ln)) + ln) ln) ln) + ln) ln 4 ). 8. Calculate the double integral sin + y) da, where [, π/] [, π/]. Solution. In this case it is convenient to integrate first with respect to the variable y. We have π/ π/ sin + y) da sin + y) dy d π/ π/ π/ cos + y) yπ/ y ) d cos + π ) ) + cos) d cos) d π/ cos + π ) d. These two single integrals can be computed easily using integration by parts, and this way we get sin + y) da π. 9. Calculate the double integral da, where [, ] [, ]. + y

6 Solution. We will need to use the identity lna + ) d lna + ) + a arctan. *) a) which can be obtained using integration by parts. Integrating first with respect to the variable, we have + y da d dy + y ln + y ) ) dy ln4 + y ) dy y ln4 + y ) y + 4 arctan d dy + y ) ln + y ) dy ) y y ) y y ln + y ) y + arctany) ) y y here we have used the identity *)) ln 5 ) + arctan ) arctan).. Find the volume of the solid that lies under the hyperbolic paraboloid z y, and above the square [, ] [, ]. Solution. We can see that the function f, y) y is nonnegative over the given rectangle. Therefore, calling S the solid we have VolS) Thus VolS) cubic units. y ) da y y ) ) dy dy. y ) d dy