# Math Help - Help with optimization of a box

1. ## Help with optimization of a box

Concrete costs $15 per square meter, and glass costs$100 per square meter. If the volume of the pool must be 1000 cubic meters, what should the dimensions be to minimize the cost of the pool? 1 of the side is made of glass and the bottom and other 3 sides are made out of concrete.

xyz = 1000
z= 1000/xy
surface area = xy +2yz + 2xz
elimate:
SA = xy + 2000/x + 2000/y
Then I take the derivative in terms of x and y
fx(x,y) = y -2000/x^2
fy(x,y) = x -2000/y^2

Then I'm lost here.

Can anyone help? Thanks

2. Let $C_0$ be the concrete cost ($15 per square meter) Let $G_0$ be the glass cost ($100 per square meter)
Let $V_0$ be the volume of the pool (1000 cubic meters)
Let z be the height of the pool

$xyz\;=\;V_0$

We try to minimize the cost C(x,y,z)

$C(x,y,z)\;=\;G_0yz + C_0(yz+2xz+xy)$

$C(x,y)\;=\;G_0\frac{V_0}{x} + C_0(\frac{V_0}{x}+2\frac{V_0}{y}+xy)$
Let us differentiate
x : $-G_0\frac{V_0}{x^2} - C_0\frac{V_0}{x^2}+C_0y\;=\;0$

y : $-2\frac{C_0V_0}{y^2} + C_0x\;=\;0$

Second equation gives $x\;=\;2\frac{V_0}{y^2}$

Replacing in first equation gives at the end
$y\;=\;(4\frac{C_0}{G_0+C_0}V_0)^{\frac{1}{3}}$

Then
y=8.05m
x=30.86m
z=4.03m

PS : I suggest to name the constant in order to be able to find an eventual calculation mistake using the units

For instance
$y\;=\;(4\frac{C_0}{G_0+C_0}V_0)^{\frac{1}{3}}$

$\frac{C_0}{G_0+C_0}$ has no unit

$V_0^{\frac{1}{3}}$ has meter as unit
Therefore the result is consistent (but it does not mean that it is true !)

3. Hello, khuezy!

Concrete costs $15/m², and glass costs$100/m².
If the volume of the pool must be 1000 m³,
what should the dimensions be to minimize the cost of the pool?
One of the sides is glass; the bottom and other 3 sides are concrete.

$xyz \:= \:1000 \quad\Rightarrow\quad z\:= \:\frac{1000}{xy}$ .[1]

Surface area: . $xy +2yz + 2xz$

This is all correct, but you forgot about the Cost.
Code:
         *---------*
/|        /|
/ |       / |z
*---------*  |
|         |  *
z|    G    | /y
|         |/
*---------*
x

Suppose the front face is glass.
Its area is $xz$ m² and costs $100/m². . . Its cost is: . $100xz$ dollars. The rest of the area is: . $xy + 2yz + xz$ m² at$15/m².
. . Its cost is: . $15(xy + 2yz + xz)$ dollars.

The total cost is: . $C \;=\;15(xy + 2yz + xz) + 100xz \;=\;15xy + 30yz + 115xz$

Substitute [1]: . $C \;=\;15xy + 30y\left(\tfrac{1000}{xy}\right) + 115x\left(\tfrac{1000}{xy}\right)$

. . and we have: . $C \;=\;15xy + 30,\!000x^{-1} + 115,\!000 y^{-1}$

Set the partial derivatives equal to zero . . .

. . $\begin{array}{cccccc}C_x &=& 15y - 30,\!000x^{-2} &=& 0 & {\color{blue}[2]} \\

C_y &=& 15x - 115,\!000y^{-2} &=& 0 & {\color{blue}[3]} \end{array}$

Multiply [2] by $x^2\!:\;\;15x^2y - 30,\!000\:=\:0 \quad\Rightarrow\quad y \:=\:\frac{2000}{x^2}$ .[4]

Multiply [3] by $y^2\!:\;\;15xy^2 -115,\!000\:=\:0 \quad\Rightarrow\quad 15xy^2 \:=\:115,\!000$ .[5]

Substitute [4] into [5]: . $15x\left(\tfrac{2000}{x^2}\right)^2 \:=\:115,\!000 \quad\Rightarrow\quad \frac{60,\!000,\!000}{x^3} \:=\:115,\!000
$

. . $x^3 \:=\:\frac{60,\!000,\!000}{115,\!000} \:=\:\frac{12,\!000}{23} \quad\Rightarrow\quad x \:=\:\sqrt[3]{\frac{12,000}{23}} \quad\Rightarrow\quad\boxed{ x \;=\;\frac{10}{23}\sqrt[3]{6348}}$

Substitute into [4]: . $y \;=\;\frac{200}{\left(\frac{10}{23}\sqrt[3]{6348}\right)^2} \quad\Rightarrow\quad\boxed{ y \;=\;\frac{5}{3}\sqrt[3]{6348}}$

Substitute into [1]: . $z \;=\;\frac{1000}{\left(\frac{10}{23}\sqrt[3]{6348}\right)\left(\frac{5}{3}\sqrt[3]{6348}\right)} \quad\Rightarrow\quad\boxed{ z \;=\;\frac{5}{23}\sqrt[3]{6348}}$