quiz7soln

# Quiz7soln - L 2 W 2 H = 108 We are going to use Lagrange multipliers here Let g L,W,H = L 2 W 2 H We ﬁnd that ∇ V = h WH,LH,LW i and ∇ g = h

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MATH 231 / Spring 2008 March 19, 2008 Name: Key 1. (2 points) State Fubini’s Theorem. Solution: If f is continuous on the rectangle R = { ( x,y ) | a x b,c y d } , then ZZ R f ( x,y ) dA = Z b a Z d c f ( x,y ) dy dx = Z d c Z b a f ( x,y ) dxdy 2. (8 points) A package in the shape of a rectangular box can be mailed by the US Postal Service if the sum of its length and girth (the perimeter of a cross-section perpendicular to the length) is at most 108 in. Find the dimensions of the package with largest volume that can be mailed. Solution: We know that the girth is equal to twice the width plus twice the height. Thus, let L , W , and H denote the length, width, and height of our box respectively. Clearly, the box with the largest volume will use all 108 in. of potential. Thus, letting V denote the volume of the box, we have the following equations: V ( L,W,H ) = LWH
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Unformatted text preview: L + 2 W + 2 H = 108 We are going to use Lagrange multipliers here. Let g ( L,W,H ) = L + 2 W + 2 H . We ﬁnd that ∇ V = h WH,LH,LW i and ∇ g = h 1 , 2 , 2 i . Computing ∇ V = λ ∇ g , we ﬁnd that we have four equations: WH = λ, LH = 2 λ, LW = 2 λ, L + 2 W + 2 H = 108 Using the second and third equations, along with the fact that the length of the box can’t be zero (that would be one lame box, after all), we ﬁnd that LH = LW , so H = W . Thus, going back to look at the ﬁrst and second equations, we know that LH = 2 λ = 2 WH = 2 H 2 , so L = 2 H . Now let’s plug all this jazz into the last equation and solve: 108 = L + 2 W + 2 H = 2 H + 2 H + 2 H = 6 H, so H = 108 / 6 = 18 Thus, W = 18 and L = 2 · 18 = 36. 1...
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## This homework help was uploaded on 04/17/2008 for the course MATH 231 taught by Professor Bociu during the Spring '08 term at UVA.

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