Maximum point question (1 Viewer)

nazfiz · Sep 30, 2011

A rectangular area is to be fenced and divided into six rectangles by one dividing fence parallel to its length and two fences parallel to its breadth. If the total length of fencing available is 120m, find the maximum possible area.

Can someone please answer this with working out
Thanks

SpiralFlex · Sep 30, 2011

First of all always draw a pretty diagram. I will let the length be $y$ and the breadth be $h$

From our pretty diagram, in our breadth section. We can see one length is $\frac{h}{2}.$ Similarly in our length section, one length is $\frac{y}{3}.$

Let's somehow related this. I know! We can use perimeter!

We have 4 breadths and only 3 lengths. So,

$4h+3y=120$

$y=40-\frac{4}{3}h$

One equation generated. We obviously need more. Let's generate an equation for the area. We know the breadth is $h$ . We just generated an equation involving $y.$ So we must know connect these two into an area format.

$A=h(40-\frac{4}{3}h)$

$=40h-\frac{4}{3}h^2$

This I am sure you know where to go from here.

$\frac{dA}{dh}=40-\frac{8}{3}h$

For maximum area or minimum area, $\frac{dA}{dh}=0$

$\therefore 40-\frac{8}{3}h=0$

$h=15$ metres. Now we must test for maximum/minimum.

Clearly from our table a maximum area will occur when $h=15$ . So we can substitute this into our area formula.

$A=(15)(40-\frac{4}{3}(15))$

$\therefore A=300\: \textup{metres}^2$

bleakarcher · Sep 30, 2011

i got 300m^2 as well

SpiralFlex · Sep 30, 2011

bleakarcher said:
i got 300m^2 as well

I must not be old after all.

nazfiz · Sep 30, 2011

Thanks guys. It was the generation of the equation part that had me stunned.

Maximum point question (1 Viewer)

nazfiz

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SpiralFlex

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bleakarcher

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SpiralFlex

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nazfiz

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