What Is the Maximum Area of This Isosceles Trapezoid?

[checkitagain]

Given:

an isosceles trapezoid

The shorter base is 8 units in length

The slant height is 12 units in length.What is the maximum area that this isosceles trapezoid could have?

You could answer with either an exact number, or an answer
where the value is to the nearest hundredths of square units,
if the exact value is "too cumbersome."

 

[CaptainBlack]

Given:

an isosceles trapezoid

The shorter base is 8 units in length

The slant height is 12 units in length.What is the maximum area that this isosceles trapezoid could have?

You could answer with either an exact number, or an answer
where the value is to the nearest hundredths of square units,
if the exact value is "too cumbersome."

Let the other base be of length \(8+2x\), then the area is:

\[ A(x)=8\sqrt{12^2-x^2}+x\sqrt{12^2-x^2} \]

subject to the constraints that \( x \in [0,12] \).

This is a constrained maximisation problem, so the maximum occurs at a calculus type local maxima of \(A(x)\) in the interior of the feasible region or at a boundary point.

The maximum area is \(\approx 146.345 \) square units and occurs when \( x=2(-1+\sqrt{19} ) \approx 6.7178 \) units

Exact area (courtesy of Maxima): \(\left( 2\,\sqrt{19}+6\right) \,\sqrt{8\,\sqrt{19}+64}\)

(note: copy and past of LaTeX from Maxima works perfectly)

CB