Question

Developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 308 feet of fencing is not fence the side along the street what is the largest area that can be enclosed

Answer 1

Here is what we know:
•Perimeter is 308 ft
•Fencing is rectangular in figure but only encloses 3 sides.


Let x=width and y=length

We can therefore form two systems of equations:
\((1)...308=2x+y\)
\((2)...A=xy\)
Now we are supposed to find the largest area enclosed using these requirements. You must first solve for y in equation 1 so that you can use equation 2 to find the max area.
\((1)...308=2x+y\)
\(308-2x=y\)
∴\(A=x(308-2x)\)
\(A=308x-2x^2\)

Now, to find the maximum, you must first find the critical points, where \(\frac{dA}{dx}=0 ~\text{and}~DNE \). Take the derivative.
\(A=308x-2x^2\)
\(\frac{dA}{dx}=308-4x\)
\(0=308-4x\)
\(x=\frac{-308}{-4}=77\)
\(\frac{1}{0}=308-4x\)
There is no such value of x that yields an indeterminate value. Hence, your critical point is x=77. Now, you must test whether this is a maximum, minimum, or neither. Let's use the first derivative test.
\(•\frac{dA}{dt}|_{t<77}\)
\(\frac{dA}{dx}|_{t=70}=28\)
\(•\frac{dA}{dx}|_{t>77}\)
\(\frac{dA}{dt}|_{t=80}=-12\)
∴Since \(\frac{dA}{dt}\) changes from positive to negative at x=77, when the width is 77, the area reaches a maximum.


But that is not the full question. We are asked to find the AREA. So since we know x=77, find y.
\(308=2(77)+y\)
\(y=154\)
\(A_{max}=(77)(154)=11,858~ft^2\)

Answer 2

Or you could just take that area function and find the vertex, it doesn't matter