Find the dimensions of the box that minimize the amount of material used, that is, the surface area of the box must be minimized

Question

A rectangular box is to be made to have a volume of 64, 000 cm3. Its base is a rectangle whose length is twice as much as its width and the box has an open top. Find the dimensions of the box that minimize the amount of material used, that is, the surface area of the box must be minimized. Note that the box has an open top meaning no material is to be used for the top.

Answer 1

x*2x*h=64000

2x^2*h= "

Here's the formula; do your substituting.

There is several combinations:

I first used 10 and 320 = 64000;

mesurments: =10*20*320

so 10*320*2+20*32*2+10*20=7880sq/cm

This is to be done on any combination until you find a smaller sq/cm.

Answer 2

Without a top, the surface area of a box with dimensions x, y, and z is:

S = xy + 2yz + 2xz = xy + 2z(x + y)

Since the length is twice as much as the width, we have y = 2x, therefore we can rewrite the above equation as:

S = 2x² + 2z(3x) = 2x² + 6xz

We also know that the volume is 64000 cm³, meaning:

xyz = 64000

2x²z = 64000

z = 32000/x²

We can finally rewrite our surface area in terms of one variable as:

S(x) = 2x² + 192000/x²

Now, to maximize S, we must find x such that dS/dx(x) = 0. We have:

S'(x) = 4x - 384000/x³ = 0

4x = 384000/x³

4x⁴ = 384000

x⁴ = 96000

x (width) = 17.6022 cm

y (length) = 2x = 35.2044 cm

z (height) = 32000/x² = 103.2800 cm