1. ## 2 Optimization Problems

An open topped rectangular box with a square base is to be constructed with a volume of $32m^{3}$. Find the Dimension that require the lease amount of surface material.

and...

In microcomputers, most of the components are squeezed into a single box-shaped block. If the block has a length equal to twice the width and if the total surface area of the block must be $200cm^{2}$ in order to dissipate the heat produced, find the dimensions for the maximum volume of the block.

I couldn't figure out how to get equations out of these two problems.

Thanks.

2. Originally Posted by johntuan
...

In microcomputers, most of the components are squeezed into a single box-shaped block. If the block has a length equal to twice the width and if the total surface area of the block must be $200cm^{2}$ in order to dissipate the heat produced, find the dimensions for the maximum volume of the block.

...
Hello,

surface area of a rectangular block: $a = 2\cdot l\cdot w + 2\cdot l\cdot h + 2\cdot w\cdot h$

Since l = 2w and a = 200 you get:

$200 = 2\cdot 2w \cdot w + 2\cdot 2w \cdot h + 2\cdot w\cdot h~\implies~200 = 4w^2 + 6wh~\implies~h = \frac{200}{6w} - \frac{4w^2}{6w}$

The volume of a rectangular block is $V=l \cdot w \cdot h$

Plug in the values you already know:

$V(h)=2w^2 \cdot \left( \frac{200}{6w} - \frac{4w^2}{6w} \right)$

Expand the bracket and calculate the minimum using the first derivative ....and so on.

3. How would you expand the bracket?

4. Originally Posted by johntuan
How would you expand the bracket?
Hello,

multiply and cancel equal factors:

$
V(h)=2w^2 \cdot \left( \frac{200}{6w} - \frac{4w^2}{6w} \right) = \frac{200}{3} w - \frac43 w^3$
......... Therefore

$V'(h) = \frac{200}{3}-4w^2$........ To calculate the minimum V'(h) = 0:

$w = \sqrt{\frac{50}{3}}\approx 4.08...$ ........ Now plug in this value to calculate l and h.

$\left(l=\sqrt{\frac{200}{3}},~h=\frac{20}{9} \cdot \sqrt{6} \right)$