# Thread: Differentiation Maximization Word Problem.

1. ## Differentiation Maximization Word Problem.

A rectangular sheet of cardboard, 50 cm by 40 cm, is to have square corners cut of so it can be folden into a rectangular tray.

Find the maximum volume possible for the tray.

Here is what I got so far.

Volume of Tray: L x W x H
Volume of Tray with variable: L=(50 - 2x) x W(40 - 2x) x H=x
Where x is equal to the cut out squares.
(50 - 2x)(40 - 2x)x
(2000 - 100x - 80x +4x^2)x
2000x - 100x^2 - 80x^2 + 4x^3
F'(x) = 2000 - 40x + 12x^2
4(500 - 40x + 3x^2)

Ahaha, I'm not sure what is going wrong...

2. Originally Posted by Liparulo
A rectangular sheet of cardboard, 50 cm by 40 cm, is to have square corners cut of so it can be folden into a rectangular tray.

Find the maximum volume possible for the tray.

Here is what I got so far.

Volume of Tray: L x W x H
Volume of Tray with variable: L=(50 - 2x) x W(40 - 2x) x H=x
Where x is equal to the cut out squares.
(50 - 2x)(40 - 2x)x
(2000 - 100x - 80x +4x^2)x
2000x - 100x^2 - 80x^2 + 4x^3
F'(x) = 2000 - 40x + 12x^2
4(500 - 40x + 3x^2)

Ahaha, I'm not sure what is going wrong...
Looks good so far.

You have $\displaystyle V = (50 - 2x)(40 - 2x)x$

$\displaystyle = 2000x - 180x^2 + 4x^3$.

To find the maximum, you have to differentiate, set equal to 0 and solve for $\displaystyle x$.

$\displaystyle \frac{dV}{dx} = 2000 - 360x + 12x^2$

$\displaystyle 12x^2 - 360x + 2000 = 0$

Solve for $\displaystyle x$.