can u help me start this?

Question

A builder intends to construct a storage shed having a volume of 900 ft^3. a flaf roof, and a rectangular base whose width is three-fourths the length. The cost per square foot of the materials is $4 for the floor, $6 for the sides, and $3 for the roof. What dimensions will minimize the cost?

Answer 1

Let x = length
3/4 x = width
900/(3/4 x^2) = 1200/x^2 = height

floor = 4(3/4 x^2) = 3x^2
sides = 6{2[x(1200/x^2)] + 2[3/4 x(1200/x^2)]}
= 25200/x
roof = 3(3/4 x^2) = 9/4 x^2

C(x) = 3x^2 + 25200/x + 9/4 x^2
= 21/4 x^2 + 25200/x

Okay, that's your cost function. Take it's derivative and set to 0. Your solution will give you the x that will minimize cost.

Phew!

Answer 2

Try to set up parameters:
x y and z are the length, width, and height respectively.

We know that xyz = 900.
We also know that y = 3/4 x
We finally know that z = 900/xy from equation 1 = 1200/x^2

Therefore, the cost = 4(xy) + 5(2xz+2yx) + 3(xy), where 4 5 and 3 are the prices per square foot of the floor, sides, and roof.

I want everything in terms of one variable, so I replace all the y's and z's with x's using the expressions I wrote above. Take the derivative of the cost equation with respect to x and set it equal to zero. Solve for x, and from there, you can get y and z using those same expressions. This is a lot of work for me to type out, but you can probably take over from here without someone doing all the explicit work for you =).

Answer 3

L*w*h = 900
w = .75L
So, .75hL^2 = 900
hL^2 = 1200
h = 1200/L^2
minimize 4Lw + 6*2(L+w)h + 3Lw
7Lw +12hL + 12hw
7L(.75L) + 12hL + 12h(.75L)
5.25L^2 + 12hL + 9hL (substituting for w)
5.25L^2 + 21hL
5.25L^2 + 21(1200/L^2)L =
minimize 5.25L^2 + 25200/L
minimize L^2 + 4800/L