Question

peter has 400 yards of fencing to enclose a rectangular area. find the dimensions of the rectangle that maximize the enclosed area. what is the maximum area? a rectangle that maximizes the enclosed area has a length of yards and a width of yards.

Explanation:

Step1: Set up the perimeter and area equations

Let the length of the rectangle be $l$ and the width be $w$. The perimeter $P = 2l + 2w$. Given $P=400$, so $2l + 2w=400$, which simplifies to $l + w = 200$, and $l=200 - w$. The area $A=l\times w=(200 - w)w=200w - w^{2}$.

Step2: Find the maximum of the area function

The area function $A(w)=200w - w^{2}$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=-1$, $b = 200$, $c = 0$. The vertex of a quadratic function $y = ax^{2}+bx + c$ has its $x$ - coordinate (in our case $w$ - coordinate) at $w=-\frac{b}{2a}$. Substituting $a=-1$ and $b = 200$ into $w=-\frac{b}{2a}$, we get $w =-\frac{200}{2\times(-1)} = 100$.

Step3: Find the length

Since $l=200 - w$, when $w = 100$, $l=200-100 = 100$.

Step4: Calculate the maximum area

Substitute $l = 100$ and $w = 100$ into the area formula $A=l\times w$. So $A=100\times100 = 10000$ square - yards.

Answer:

A rectangle that maximizes the enclosed area has a length of 100 yards and a width of 100 yards. The maximum area is 10000 square yards.