nice cal question 1

[pudder]

a box of constant volume (c) is to be twice as long as its width. the material on the top and four sides cost three times as much per square meter as the on the bottom. what are the most economical dimensions.

[pudder]

can anyone help solve this problem
thanks

[OfficeShredder]

Call the dimensions h for height, w for width, and d for depth.

We know h*w*d=C, a constant (the volume).

You can also develop equations for the surface area of the cheap sides, and the expensive side. So the total cost is 3*surface area of expensive side + surface area of cheap sides. Then, the question is how do you minimize that cost?

Unfortunately, I just realized my terminology doesn't line up well with the problem. So let's call the depth the length, and let that be the "big" dimension (d=2*w)

[HallsofIvy]

Let x be the width. Then, since the length is twice the width, the length is 2x. Let h be the height. Then the volume of the box is "width time length times height"= (2x)(x)(h)= 2x^2 h= c.

The bottom has width x and length 2x and so area 2x^2. Taking the cost of unit area of the other sides to be 1, the bottom costs 3(2x^2)= 6x^2.
Two of the sides have width x and height h and so area and cost xh: the two together cost 2xh. The other two sides have width 2x and height h and so area and cost 2xh: the two together cost 4xh. Finally, the top has width x and length 2x and so area 2x^2 and so cost 2x^2. The total cost of the box is 2xh+ 4xh+ 2x^2+ 6x^2= 8x^2+ 6xh. The problem, then, is to minimize 8x^2+ 6xh subject to the constraint that 2x^2h=c. From that constraint,
h= c/2x^2. Putting that into the object function, we need to minimize 8x^2+ 6x(c/2x^2)= 8x^2+ 3c/x= 8x^2+ 3cx^-1. The derivative of that is
16x- 3cx^-2. Set the derivative equal to 0. 16x- 3cx^-2= 0 so 16x= 3cx^-2. Multiplying by x^-2, 16x^3= 3c or x^3= 3c/16. the width, x, is the cuberoot of 3c/16. The length, 2x, is twice that. The height, h, can then be found from h= c/2x^2.