# Thread: optimization problem

1. ## optimization problem

A cardboard box without a lid is to have a volume of 32,000 cm^3. Find the dimensions that minimize the amount of cardboard used.

2. Hello, jlt1209!

This requires partial derivatives . . .

A cardboard box without a lid is to have a volume of 32,000 cm³.
Find the dimensions that minimize the amount of cardboard used.
Code:
         *- - - -*
/|      /|
/ |     / | z
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z |       | / y
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x

The length, width, height of the box are: . $x,y,z$, respectively.

The volume is 32,000 cm³: . $xyz \:=\:32,\!000 \quad\Rightarrow\quad z \:=\:\frac{32,\!000}{xy}$ .[1]

The total surface area of the box is: . $A \;=\;xy + 2xz + 2yz$ .[2]

Substitute [1] into [2]: . $A \;=\;xy + 2x\left(\frac{32,\!000}{xy}\right) + 2y\left(\frac{32,\!000}{xy}\right)$

. . and we have: . $A \;=\;xy + 64,\!000y^{-1} + 64,\!000x^{-1}$

Set the partial derivatives equal to 0.

. . $\begin{array}{cccc}
\dfrac{\partial A}{\partial x} \;=\;y - 64,\!000x^{-2} \;=\;0 & \Longrightarrow & y \:=\:\dfrac{64,\!000}{x^2} & {\color{blue}[3]}\\ \\[-3mm] \dfrac{\partial A}{\partial y} \;=\;x - 64,\!000y^{-2} \;=\;0 & \Longrightarrow & x \:=\:\dfrac{64,\!000}{y^2} & {\color{blue}[4]}
\end{array}$

Substitute [3] into [4]: . $x \:=\:\frac{64,\!000}{\frac{64,000^2}{x^4}} \quad\Rightarrow\quad x \:=\:\frac{x^4}{64,\!000}$

. . $x^4 - 64,\!000x\:=\:0 \quad\Rightarrow\quad x(x^3-64,\!000) \:=\:0$

. . $x^3 \:=\:64,\!000 \quad\Rightarrow\quad\boxed{ x \:=\:40}$

Substitute into [3]: . $y \:=\:\frac{64,\!000}{40^2} \quad\Rightarrow\quad\boxed{ y \:=\:40}$

Substitute into [1]: . $z \:=\:\frac{32,\!000}{40\cdot40} \quad\Rightarrow\quad\boxed{ z \:=\:20}$

Therefore: . $\begin{Bmatrix}\text{Length} &=& 40\text{ cm} \\ \text{Width} &=& 40\text{ cm} \\ \text{Height} &=& 20\text{ cm} \end{Bmatrix}$

3. Originally Posted by jlt1209
A cardboard box without a lid is to have a volume of 32,000 cm^3. Find the dimensions that minimize the amount of cardboard used.
Are you familiar with the method of Lagrange multipliers?

$L = xy + 2yz + 2zx + \lambda (xyz - 32, 000)$ and ppplication of this method leads to the following equations that need to be solved simultaneously:

$0 = y + 2z + \lambda y z$ .... (1)

$0 = x + 2z + \lambda x z$ .... (2)

$0 = 2y + 2x + \lambda x y$ .... (3)

$32,000 = xyz$ .... (4)