2
$\begingroup$

TX farmer has 100 metres of fencing to use to make a rectangular enclosure for sheep as shown.

He will use existing walls for two sides of the enclosure and leave an opening of 2 metres for a gate.schema of enclosure

a) Show that the area of the enclosure is given by: $A = 102x – x^2.$

b) Find the value of x that will give the maximum possible area.

c) Calculate the maximum possible area.

How do I assign the two variables for area ? Can anyone assist me in solving this problem?

$\endgroup$
1
$\begingroup$

From your picture, one side of the rectangle is x.
Since you have 100 metres, this means the other side has length (100 - x) + 2 = 102 - x. So, the Area A = $(102 - x) * x = 102x - x^2$

The maximum area occurs where $\frac{dA}{dx} = 102 - 2x = 0$
or where $x = 51$

So, the max area A = $102(51) - 51^2$

$\endgroup$
  • $\begingroup$ You're welcome. $\endgroup$ – user137481 Mar 17 '15 at 21:06
  • $\begingroup$ thank you for your assistance and guiding me in the solution. Area=length x width Width = x Length of the enclosure parallel to the wall = (100-x) +2 =102 - x Area= x(102-x) Area, A = 102x-x^2 dA/dx=102- 2x At Stationary Point dA/dx=0 ⇒ 102 – 2x = 0 2x=102 x=51 the maximum area occurs when x = 51 and Area, A= (102x-x^2) = 102x51 – (51)^2 = 2600 $\endgroup$ – MsK05 Mar 17 '15 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.