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Question
diana has 1600 yards of fencing and wishes to enclose a rectangular area. (a) express the area a of the rectangle as a function of the width w of the rectangle. (b) for what value of w is the area largest? (c) what is the maximum area? (a) a(w) =
Step1: Find the length in terms of width
The perimeter of a rectangle is $P = 2L+2W$. Given $P = 1600$, so $2L + 2W=1600$. Solving for $L$, we get $L=\frac{1600 - 2W}{2}=800 - W$.
Step2: Express the area as a function of width
The area of a rectangle is $A = L\times W$. Substituting $L = 800 - W$ into the area formula, we have $A(W)=(800 - W)W=800W - W^{2}$.
Step3: Find the maximum of the area function
The function $A(W)=-W^{2}+800W$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=-1$, $b = 800$ and $c = 0$. The vertex of a quadratic function $y=ax^{2}+bx + c$ has its $x$-coordinate (in our case $W$-coordinate) at $W=-\frac{b}{2a}$. So $W=-\frac{800}{2\times(-1)} = 400$.
Step4: Calculate the maximum area
Substitute $W = 400$ into the area function $A(W)$. $A(400)=800\times400-400^{2}=320000 - 160000=160000$.
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(a) $A(W)=800W - W^{2}$
(b) $W = 400$
(c) $160000$ square - yards