Answer
(a) $A(x)=1500x-x^2$
(b) 750
(c) 562,500 squared feet.
Work Step by Step
The perimeter of a rectangle is the addition of all its sides:
$P=2w+2l$
The area of a rectangle is width times length:
$A=w\cdot l$
Since we want the area as a function of $w$, we must re-write $l$ in terms of $w$. Since we know the value of the perimeter, we can solve for $l$:
$3000=2w+2l$
$3000-2w=2w+2l-2w$
$3000-2w=2l$
$(3000-2w)/2=2l/2$
$l=1500-w$
Now we can substitute $l$ and get the area as a function of its width, which we'll change to x as indicated in the exercise:
$A(x)=x\cdot (1500-x)$
$A(x)=1500x-x^2$
In order to answer (b) and (c), it would be useful to re-write the quadratic equation to vertex form since the vertex is the global maximum:
$A(x)=-(x^2-1500x)$
$A(x)=-(x^2-1500x+(\frac{1500}{2})^2)+(\frac{1500}{2})^2$
$A(x)=-(x^2-1500x+562500)+562500$
$A(x)=-(x-750)^2+562500$
So, the value of x where the area is the largest is the x-value of the vertex which is 750. The maximum area is the y-value of the vertex which is 562,500 squared feet.
