Derivative Applications Question (High-school Calculus)

[NDiggity]

1. The problem statement, all variables and given/known data
A box with a square base and an open top has a volume of 4500cm^3. What are the dimensions of the box that will minimize the amount of material used? (Remember that the amount of material used refers to the surface area of the box).

My buddy and I spent a solid hour on this questions with no luck. Please help!

2. Relevant equations
No set equations, make your own.

3. The attempt at a solution

First we set our variables:
Let x=a side of the base
Let y=height of the box

Then we isolated the y variable:
x*x*y=4500
x^2*y=4500
y=4500/x^2

Then we made a surface area equation:
S(x)=x^2 + 4xy
S(x)=x^2 + 4x(4500/x^2)
S(x)=x^2 + 18000x/x^2
Then we multiplied everything by x^2 to get rid of the denominator to get:
S(x)=x^4 + 18000x
S(x)=x(x^3 + 18000)

Then we found the derivative of this equation:
S'(x)=4x^3 + 18000
S'(x)=4(x^3 + 4500)

This is where we got stuck because the critical number will be negative. Please help, we do not know what we did wrong.

[FunkyDwarf]

Then we multiplied everything by x^2 to get rid of the denominator to get: If you're going to do that you need to multiply S(x) as well as you are not multiplying by one.

[Doc Al]

Then we isolated the y variable:
x*x*y=4500
x^2*y=4500
y=4500/x^2

Then we made a surface area equation:
S(x)=x^2 + 4xy
S(x)=x^2 + 4x(4500/x^2)
S(x)=x^2 + 18000x/x^2
So far, so good. Simplify this and find its minimum.

Then we multiplied everything by x^2 to get rid of the denominator to get:
S(x)=x^4 + 18000x
S(x)=x(x^3 + 18000)
Say what now? :yuck: You can't just arbitrarily multiply by x^2! For one thing, x^2*S(x) does not equal S(x).

[NDiggity]

Ok so if I don't multiply by x^2 and simplify, I'm left with:
S(x)=x^2 + 18000/x
S(x)=x^2 + 18000x^-1

So with the derivative I get:
S'(x)=2x -18000x^-2
S'(x)=2x -18000/x^2

Is this better?

[Doc Al]

Much better. :smile:

[NDiggity]

Yay, thank you very much for the help so far! So, would my critical number be the cube root of 9000? If so my dimensions are cube root of 9000 or 20.8 cm by 10.4cm.

[Doc Al]

Looks good to me. :smile:

[esalihm]

yeah correct