# Find the Max Volume of a box using Calc

• May 19th 2011, 05:55 AM
MrJoe2000
Find the Max Volume of a box using Calc
Find the max volume of a box with a square base if you have 1200 sq. ft. of material for sides and a bottom (assuming open top).

I'm screwing up somewhere but I can't figure out where. V= x^2*y and then your y= (1200-x^2)/4x

Using a graphical method I found the answer x to be 20. But how do you get here using calculus?
• May 19th 2011, 06:11 AM
Mondreus
$\displaystyle V=x^2y=x^2(\frac{1200-x^2}{4x}=\frac{1200x-x^3}{4}$

Find where:
$\displaystyle V'(x)=0$
• May 19th 2011, 06:23 AM
MrJoe2000
Okay, I see that you substituted the y equation immediately into the volume equation.

Why couldn't you take the derivative of x^2*y to get:

vol'(x)= x^2((1200-x^2)(4) - 4x(-2x))/(4x)^2 + 2xy

as in lets v=X^2 and let u=y

then vol'(x)= v*u' + u*v'

When I work this out I get x=60 and y= -10 which is obviously NOT the answer
• May 19th 2011, 09:08 AM
HallsofIvy
Quote:

Originally Posted by MrJoe2000
Okay, I see that you substituted the y equation immediately into the volume equation.

Why couldn't you take the derivative of x^2*y to get:

vol'(x)= x^2((1200-x^2)(4) - 4x(-2x))/(4x)^2 + 2xy

You are differentiating y wrong, using the quotient rule backwards:
dy/dx= (-2x(4x)- 2x(1200- x^2)/(4x)^2

Quote:

as in lets v=X^2 and let u=y

then vol'(x)= v*u' + u*v'

When I work this out I get x=60 and y= -10 which is obviously NOT the answer
• May 19th 2011, 09:14 AM
MrJoe2000
Why do I take the quotient rule backwards?