# dimensions of box optimization

• Oct 29th 2008, 02:55 PM
iwonder
dimensions of box optimization
how to find the dimensions of the rectangular box having the largest volume and surface area 150 square units?
• Oct 29th 2008, 03:14 PM
skeeter
completely closed box or an open top box?
• Oct 29th 2008, 04:11 PM
iwonder
completely closed...thankyou
• Oct 29th 2008, 05:04 PM
skeeter
I was afraid you'd say that.

The max volume for a closed box with a fixed surface area will be a cube.

Here is the work ...

let x , y , and z be the length, width, and height.

$\displaystyle V = xyz$

$\displaystyle A = 150 = 2xy + 2xz + 2yz$

$\displaystyle xy + xz + yz = 75$

$\displaystyle z = \frac{75-xy}{x+y}$

$\displaystyle V = \frac{75xy - x^2y^2}{x+y}$

$\displaystyle \frac{\partial V}{\partial x} = \frac{(x+y)(75y - 2xy^2)-(75xy - x^2y^2)(1)}{(x+y)^2}$

$\displaystyle \frac{\partial V}{\partial x} = \frac{75xy - 2x^2y^2 -2xy^3 + 75y^2 - 75xy + x^2y^2}{(x+y)^2}$

$\displaystyle \frac{\partial V}{\partial x} = \frac{75y^2-x^2y^2-2xy^3}{(x+y)^2}$

since $\displaystyle y \neq 0$

$\displaystyle \frac{\partial V}{\partial x} = \frac{75-2xy-x^2}{(x+y)^2}$

$\displaystyle \frac{\partial V}{\partial y} = \frac{(x+y)(75x - 2x^2y)-(75xy - x^2y^2)(1)}{(x+y)^2}$

$\displaystyle \frac{\partial V}{\partial y} = \frac{75x^2 - 2x^3y + 75xy - 2x^2y^2 - 75xy + x^2y^2}{(x+y)^2}$

$\displaystyle \frac{\partial V}{\partial y} = \frac{75x^2 - 2x^3y - x^2y^2}{(x+y)^2}$

since $\displaystyle x \neq 0$ ...

$\displaystyle \frac{\partial V}{\partial y} = \frac{75 - 2xy - y^2}{(x+y)^2}$

set both partials = 0 ...

$\displaystyle 75-2xy-x^2 = 0$

$\displaystyle 75 - 2xy - y^2 = 0$

combine both equations ...

$\displaystyle x^2 - y^2 = 0$

$\displaystyle (x-y)(x+y) = 0$

since $\displaystyle x \neq -y$ ... $\displaystyle x = y$

$\displaystyle z = \frac{75-xy}{x+y} = \frac{75-x^2}{2x}$

$\displaystyle V = x^2z = \frac{75x - x^3}{2}$

$\displaystyle \frac{dV}{dx} = \frac{75 - 3x^2}{2} = 0$

$\displaystyle x = 5$

$\displaystyle y = 5$

$\displaystyle z = 5$

the "cube"
• Oct 29th 2008, 05:14 PM
iwonder
thankyou very much, i really appreciate it!!!