2 Optimization Problems

**johntuan** · Jan 20th 2008, 09:25 AM

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.

**earboth** · Jan 20th 2008, 10:23 AM

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.

**johntuan** · Jan 20th 2008, 11:17 AM

How would you expand the bracket?

**earboth** · Jan 20th 2008, 09:04 PM

Originally Posted by johntuan

How would you expand the bracket?

Hello,

multiply and cancel equal factors:

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

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

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