A rectangle is to be inscribed in a right‑angled triangle whose base is $12,\text{cm}$ and height is $8,\text{cm}$. The rectangle’s base lies on the triangle’s base and its top side is parallel to that base. Find the dimensions of the rectangle that give the maximum area.

Step-by-step solution

1. Let $x$ be the width of the rectangle (along the base) and $y$ its height. Because of similar triangles, the ratio $y/(8)=(12-x)/12$ gives $y=8-\frac{2}{3}x$.
2. Express the area as a function of $x$: $A(x)=x,y=x!\left(8-\frac{2}{3}x\right)=8x-\frac{2}{3}{x}^2$.
3. Differentiate to find critical points: $A^{\prime }(x)=8-\frac{4}{3}x$. Set $A^{\prime }(x)=0$ to get $x=6$.
4. Check that this gives a maximum by noting $A^{\prime \prime }(x)=-\frac{4}{3}<0$.
5. Compute $y$ at $x=6$: $y=8-\frac{2}{3}\cdot 6=8-4=4$.
6. Thus the rectangle of maximum area has base $6,\text{cm}$, height $4,\text{cm}$, and area $24,{\text{cm}}^2$.

Concept tested: Optimization of area in a triangle

Published 2 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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