# Thread: dimensions of box optimization

1. ## dimensions of box optimization

how to find the dimensions of the rectangular box having the largest volume and surface area 150 square units?

2. completely closed box or an open top box?

3. completely closed...thankyou

4. 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.

$V = xyz$

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

$xy + xz + yz = 75$

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

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

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

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

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

since $y \neq 0$

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

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

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

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

since $x \neq 0$ ...

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

set both partials = 0 ...

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

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

combine both equations ...

$x^2 - y^2 = 0$

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

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

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

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

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

$x = 5$

$y = 5$

$z = 5$

the "cube"

5. thankyou very much, i really appreciate it!!!