Fencing a Rectangular Garden
A homeowner has 40 meters of fencing to enclose a rectangular garden against an existing straight wall, so only three sides need fencing (two widths and one length). She wants the enclosed area to be at least 150 square meters and also wants the garden to be as large as possible. She is unsure whether the maximum-area rectangle will also satisfy the area constraint, and she worries that making the garden too long will leave too little width.
1. Express the area as a quadratic in one variable and find the dimensions that maximize area.
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Let width be $w$ (two sides) and length $L$ (one side against the wall). Fencing: $2w + L = 40 \Rightarrow L = 40 - 2w$. Area: $A(w) = w(40 - 2w) = -2w^2 + 40w$. This is a downward parabola with vertex at $w = -b/(2a) = -40/(-4) = 10$, giving $L = 20$ and $A = 200$ m². So the maximum area is 200 m² at 10 m by 20 m.
2. Solve the inequality for the widths that give area at least 150 m², and interpret.
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$-2w^2 + 40w \ge 150 \Rightarrow 2w^2 - 40w + 150 \le 0 \Rightarrow w^2 - 20w + 75 \le 0 \Rightarrow (w - 5)(w - 15) \le 0$. So $5 \le w \le 15$. Widths between 5 m and 15 m meet the area goal; the maximum-area width (10 m) lies inside this interval, so both goals are compatible.
3. The homeowner insists on a length of exactly 30 m. What happens to the area constraint?
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If $L = 30$, then $2w + 30 = 40 \Rightarrow w = 5$, giving $A = 150$ m², which just meets the minimum. Any longer length forces $w < 5$ and area drops below 150. So 30 m is the longest permissible length under the constraint.