Solve the problem.
-A rectangular enclosure must have an area of at least $5400 \mathrm { yd } 2$. If $300 \mathrm { yd }$ of fencing is to be used, and the width cannot exceed the length, within what limits must the width of the enclosure lie?
A) $60 \leq w \leq 90$
B) $60 \leq w \leq 75$
C) $0 \leq w \leq 60$
D) $75 \leq w \leq 90$

Question

Quizplus · Accepted Answer

Given the perimeter (P) is 300 yards, we can express the perimeter of a rectangle as $P = 2l + 2w$, where $l$ is the length and $w$ is the width. Substituting 300 for $P$, we get $300 = 2l + 2w$, which simplifies to $150 = l + w$. The area (A) of a rectangle is given by $A = lw$, and we're told it must be at least $5400 \mathrm{yd}^2$. Since the width cannot exceed the length, $w \leq l$. From $150 = l + w$, we can express $l$ as $l = 150 - w$. Substituting into the area equation gives $5400 \leq w(150 - w)$. This simplifies to a quadratic equation $5400 \leq 150w - w^2$, or $0 \leq w^2 - 150w + 5400$. Solving this quadratic inequality for $w$, we find the feasible width values that satisfy both the area and perimeter constraints. The roots of the corresponding equation $w^2 - 150w + 5400 = 0$ are $w = 60$ and $w = 90$, but since the width must be less than or equal to the length and the total must be 150, the width must be between 60 and 75 to satisfy both the area requirement and the perimeter constraint, making the correct choice $60 \leq w \leq 75$.

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