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Question:

A 5,000 m² rectangular area of a field is to be enclosed by a fence, with a movable inner fence built across the narrow part of the field.The perimeter fence costs $10/m and the inner fence costs $4/m. Determine the dimensions of the field to minimize the cost to construct the fencing.

Is t his How I solve?:
A=5000
5000=x*w
Costs are x=10*2i y=10*2i + 4i
           =20i    =24i 
5000=20i*24i

Am I on the right track?

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  • $\begingroup$ You should define your variables. Lower case l is a poor choice because it resembles $1$. The $5000$ is an area, but the third equation right side has units of dollars, so cannot be right. $\endgroup$ – Ross Millikan Jun 15 '17 at 15:23
  • $\begingroup$ yep area equation should be $5000=lw$ $\endgroup$ – zwim Jun 15 '17 at 15:24
  • $\begingroup$ Ok so I made 5000=x*y (x width and y length) costs are x=20i y=24i i is meters.? $\endgroup$ – Guest Man Jun 15 '17 at 15:28
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Why equate the area with cost? You need to form a cost function and then optimize it.

$ Area = a \times b = 5000 \Rightarrow a = \frac{5000}{b} $

Let b be the narrow side. Then the cost function z can be written as

$z = 10(2a+2b)+4b = 20a + 24b = \frac{20 \times 5000}{b} + 24b$

Differentiating z with respect to b and equating it into zero,

$a = 77.46m \quad \& \quad b = 64.55m$

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Area $\mathcal A=5000=lw$

Since you assumed here that $w>l$ the price is $P=10\times(2l+2w)+4\times(l)$

When you substitute $w$ you get $P=24l+20w=24l+20\times\frac {5000}l$

You want to minimize the price, so let's calculate $\displaystyle P'(l)=24-\frac{100000}{l^2}$

$P'(l)=0\iff l^2=\frac{100000}{24}$ gives you $\quad l\simeq64.55\; m\quad w\simeq77.46\; m$

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  • $\begingroup$ Isn't it supposed to be 100'000 instead of 10'000? $\endgroup$ – Guest Man Jun 15 '17 at 15:58
  • $\begingroup$ you are right, silly error, I have corrected. $\endgroup$ – zwim Jun 15 '17 at 16:07
  • $\begingroup$ Thanks a lot! That's what I got (: $\endgroup$ – Guest Man Jun 15 '17 at 16:08

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