# Math Help - Applied Minimum

1. ## Applied Minimum

A container manufacturer is designing a retagular box, open at the top with a square base, that is to have a volume of 32 ft^3. If the box is to require the least amount of materials, what would be its dimensions?

I know that the answer to this equation is 4X4X2 but I'm not sure what equation to use to get this answer. I started by doing...

32= L*W*H
32= 2y*X
y=32-X/2

Am I on the right track or is there another equation I am suppost to use. If anyone know the answer could you please let me know. Thank you so much.

2. Originally Posted by johnkw87
A container manufacturer is designing a retagular box, open at the top with a square base, that is to have a volume of 32 ft^3. If the box is to require the least amount of materials, what would be its dimensions?

I know that the answer to this equation is 4X4X2 but I'm not sure what equation to use to get this answer. I started by doing...

32= L*W*H
32= 2y*X ... no
y=32-X/2
square base with side length x, height of box h ...

$32 = x^2h$

surface area ...

$A = x^2 + 4xh$

using the volume formula, solve for h in terms of x and substutute the result into the surface area equation ... find dA/dx and minimize