Solve the rational inequality. Write the solution set in interval notation.
-A rectangular enclosure must have an area of at least $800 \mathrm { yd } ^ { 2 }$ . If 180 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) $10 \leq w \leq 80$
B) $0 \leq w \leq 10$
C) $45 \leq w \leq 80$
D) $10 \leq w \leq 45$

Question

Solve the rational inequality. Write the solution set in interval notation.
-A rectangular enclosure must have an area of at least $800 \mathrm { yd } ^ { 2 }$ . If 180 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) $10 \leq w \leq 80$
B) $0 \leq w \leq 10$
C) $45 \leq w \leq 80$
D) $10 \leq w \leq 45$

Quizplus · Accepted Answer

Given the perimeter is 180 yd, the equation for the perimeter of a rectangle is $2l + 2w = 180$, where $l$ is the length and $w$ is the width. Solving for $l$, we get $l = 90 - w$. The area $A$ of the rectangle is given by $A = lw \geq 800$. Substituting $l$ gives $w(90 - w) \geq 800$. Simplifying, we get a quadratic inequality $w^2 - 90w + 800 \leq 0$. Solving this inequality gives the range for $w$ as $10 \leq w \leq 80$. However, since the width cannot exceed the length, and the length $l = 90 - w$, the maximum width $w$ can be is when $w = l$, which gives $w = 45$. Therefore, the width must lie between 10 and 45 yards, inclusive.

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