Question
"consider a rectangular industrial warehouse consisting of three seperate spaces of equal size. assume that a wall of material costs 200$ per linear ft. and the company allocates 2,400,00 for the project. 1. which dimensions maximize the total area of the warehouse 2. what is the area of each compartment in this case?
"consider a rectangular industrial warehouse consisting of three seperate spaces of equal size. assume that a wall of material costs 200$ per linear ft. and the company allocates 2,400,00 for the project. 1. which dimensions maximize the total area of the warehouse 2. what is the area of each compartment in this case?
Answer
Let each space be L*W. There are three of them, so A=3L*W The perimeter of each is 2L+2W, so the total perimeter is 6L+6W. The total number of feet is given total cost divided by cost per foot or 2,400,000/200=1200 ft. 6L+6W=1200 L+W=200 L=200-W Now you have one variable in terms of the other, A=3L*W A=3(200-W)W=600W-3W^2 . . . Differentiate A wrt W and set the derivative equal to zero to solve for the extrema. dA/dW=600-6W=0 600-6W=0 W=100 L=200-W=100 . . . You can verify using the second derivative rule that this extrema is actually a maximum. 1. L=100, W=100 maximizes the area. 2. A=L*W=10000 sq. ft.
Let each space be L*W. There are three of them, so A=3L*W The perimeter of each is 2L+2W, so the total perimeter is 6L+6W. The total number of feet is given total cost divided by cost per foot or 2,400,000/200=1200 ft. 6L+6W=1200 L+W=200 L=200-W Now you have one variable in terms of the other, A=3L*W A=3(200-W)W=600W-3W^2 . . . Differentiate A wrt W and set the derivative equal to zero to solve for the extrema. dA/dW=600-6W=0 600-6W=0 W=100 L=200-W=100 . . . You can verify using the second derivative rule that this extrema is actually a maximum. 1. L=100, W=100 maximizes the area. 2. A=L*W=10000 sq. ft.
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