An isosceles trapezoid has a shorter base of 5, a longer base of 11 and a height of 6. A rectangle is inscribed in it with one side on the longer base and its sides parallel to the bases. What is the largest possible area of the rectangle?

Step-by-step solution

1. 1

   Find how the trapezoid's width shrinks with height. The bases differ by $11-5=6$, and by symmetry each slanted side moves inward by $\tfrac{6}{2}=3$ over the full height of $6$. So the width is a linear function of the height $x$ above the bottom, dropping from $11$ at $x=0$ to $5$ at $x=6$:

   $w(x)=11-\frac{11-5}{6}\,x=11-x,\qquad 0\le x\le 6.$

2. 2

   Express the rectangle's area with one variable. Let the rectangle have height $x$, with its base on the longer base. Its top edge sits at height $x$, and it must fit inside the trapezoid there, where the figure is narrowest — the width is decreasing, so the top edge is the binding constraint. The widest such rectangle therefore has width $w(x)=11-x$ and area

   $A(x)=x(11-x)=11x-x^{2}.$

3. 3

   Maximise the quadratic. $A(x)=-x^{2}+11x$ is a downward parabola, so its maximum is at the vertex:

   $x=\frac{-b}{2a}=\frac{-11}{2(-1)}=\frac{11}{2}=5.5.$

   This value lies inside the allowed range $0\le x\le 6$, so it is a genuine maximum and not cut off by the endpoints. (Completing the square gives the same thing: $A=\tfrac{121}{4}-\left(x-\tfrac{11}{2}\right)^{2}$.)

4. 4

   Compute the maximum area.

   $A\!\left(\frac{11}{2}\right)=\frac{11}{2}\left(11-\frac{11}{2}\right)=\frac{11}{2}\cdot\frac{11}{2}=\frac{121}{4}=30.25.$

   The optimal rectangle happens to be a square of side $5.5$.

5. 5

   Sanity-check the answer. The trapezoid itself has area $\tfrac{5+11}{2}\cdot 6=48$, so the best rectangle fills $\tfrac{30.25}{48}\approx 63\%$ of it — plausible. Scanning $A(x)=x(11-x)$ numerically over $[0,6]$ in steps of $10^{-5}$ returns a maximum of $30.25$ at $x=5.5$, confirming the vertex calculation.