# A parking lot is to be formed by fencing in a rectangular plot of land except for an entrance 12 m wide. How do you find the dimensions of the lot of greatest area if 300 m of fencing is to be used?

Jan 8, 2016

The maximum area is obtained if the plot is $78$m square.

#### Explanation:

The perimeter of the plot must be $\left(300 + 12\right)$ if $300$ m fencing are used and there is a $12$ m entrance. The perimeter is given by $P = 2 \left(L + W\right)$ and the area is given by $A = L \cdot W$
$P = 2 \left(L + W\right) = 312$
$\therefore W = \frac{312}{2} - L = 156 - L$
$\therefore A = L \cdot \left(156 - L\right) = 156 L - {L}^{2}$

For a quadratic of the form $y = a {x}^{2} + b x + c$ the vertex )which gives the maximum or minimum value) is found at the point $x = c - {b}^{2} / \left(4 a\right)$
For this parking lot $c = 0 , b = 156$ and $a = - 1$
$\therefore {A}_{v} = - {156}^{2} / \left(- 4\right) = \frac{24336}{4} = 6084$

So the maximum area of the plot is 6084 sq .m.
$6084 = - {L}^{2} + 156 L$
$0 = {L}^{2} - 156 L + 6084$
$0 = {\left(L - 78\right)}^{2}$

$\therefore L = 78$
$W = 156 - l = 156 - 78 = 78$

Therefore the are of the plot is maximized if the plot is $78 m$ square.