Question

A rectangular area is to be enclosed by a wall on one side and fencing on the other three sides. If 18 meters of fencing are used, what is the maximum area that can be enclosed?

Answer 1

Answer:

$A = L W= 9*\frac{9}{2}=\frac{81}{2} m^2$

Step-by-step explanation:

For this case we assume that the total perimeter is 18 ft, we have a wall and the two sides perpendicular to the wall measure x units each one so then the side above measure P-2x= 18-2x.

And we are interested about the maximum area.

For this case since we have a recatangular area we know that the area is given by:

$A= LW$

Where L is the length and W the width, if we replace from the values on the figure we got:

$A(x)= x *(18-2x) = 18x -2x^2$

And as we can see we have a quadratic function for the area, in order to maximize this function we can use derivates.

If we find the first derivate respect to x we got:

$\frac{dA}{dx} = 18-4x=0$

We set this equal to 0 in order to find the critical points and for this case we got:

$18-4x=0$

And if we solve for x we got:

$x=\frac{18}{4}=\frac{9}{2} m$

We can calculate the second derivate for A(x) and we got:

$\frac{d^2 A}{dx^2}= -4$

And since the second derivate is negative then the value for x would represent a maximum.

Then since we have the value for x we can solve for the other side like this:

$L= 18-2x = 18-2 \frac{9}{2}= 18-9 =9m$

And then since we have the two values we can find the maximum area like this:

$A = L W= 9*\frac{9}{2}=\frac{81}{2} m^2$