ABCDE is a pentagon of fixed perimeter P cm. Its shape is such that ABE is an equilateral triangle and BCDE is a rectangle. If the length of AB is x cm, find the value of P/x for which the area of the pentagon is the maximum.
This question is in a chapter dealing with calculus and maxima and minima. Someone posted a similar question but the answer offered used LaGrange multipliers, of which I know nothing.
I have drawn the following diagram.
I have said:
Area = ABE + BCDE
$= \frac{1}{2}x(x^2 – \frac{x^2}{4})^{1/2} + xy$
$P = 2y + 3x$
So $y = \frac{P – 3x}{2}$
I now have a formula for the area in terms of P and x:
Area = $\frac{1}{2}x(x^2 – \frac{x^2}{4})^{1/2} + x(\frac {P – 3x}{2})$
I have tried differentiating this but it becomes very messy and I am not sure I am on the right track. Can anyone help?
I am self-taught.
Best Answer
$P = 3x + 2y \implies y = \frac{P - 3x}{2}$
Pentagon area $A = \frac{1}{2} \times \frac{\sqrt3 x}{2} \times x + xy$
$A = \frac{1}{2} \times \frac{\sqrt3 \, x}{2} \times x + x \times \frac{P - 3x}{2}$
At extrema, $\frac{dA}{dx} = \frac{\sqrt3 \, x}{2} + \frac{P}{2} - 3x = 0$
That gives us $\frac{P}{x} = 6 - \sqrt 3$