Question

A rectangular field will be fenced on all four sides. Fencing for the north and south sides costs $6 per foot and fencing for the other two sides costs $5 per foot. What is the maximum area that can be included for $6000

Answer 1

Answer:

The maximum area that can be included for $6000 is $A_{max}=75000 \:{ft^{2}}$

Step-by-step explanation:

We can use the fact that the area of a rectangular field is given by the formula

$A=L\cdot W$ where

$L$ = length of the field

$W$ = the width of the field

From the information given the total cost of rectangular field is

$\ 6000=2\cdot \ 6 \cdot L+2\cdot \ 5\cdot W\\\ 6000=\ 12 \cdot L+\ 10\cdot W$

From the area formula we can say $L=\frac{A}{W}$ and substitute into the total area cost as follows:

$\ 6000=\ 12 \cdot (\frac{A}{W} )+\ 10\cdot W$

And we solve for A, which is the maximum area

$\ 6000=\ 12 \cdot (\frac{A}{W} )+\ 10\cdot W\\12\cdot \frac{A}{W}+10W-10W=6000-10W\\12\cdot \frac{A}{W}=6000-10W\\\frac{12A}{W}W=6000W-10WW\\\frac{12A}{12}=\frac{6000W}{12}-\frac{10W^2}{12}\\A_{max}=\frac{6000W}{12}-\frac{10W^2}{12}$

The equation we obtain is a parabola, for this reason we need to find the W-coordinate of the vertex.

$W=-\frac{b}{2a}$

$W=-\frac{\frac{6000}{12} }{2(-\frac{10}{12} )}\\W=300$

and since

$A_{max}=\frac{6000W}{12}-\frac{10W^2}{12}$

$A_{max}=\frac{6000(300)}{12}-\frac{10(300)^2}{12}\\A_{max}=75000 \:{ft^{2}}$