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mrs_skeptik [129]
3 years ago
5

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 use

d, what is the maximum area that can be enclosed?

Mathematics
1 answer:
Rashid [163]3 years ago
8 0

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

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