# Math Help - extreme trapeze perimeter

1. ## extreme trapeze perimeter

A halfcircle with r = 14 cm has a trapeze drawn inside it.

What is the largest possible perimeter of the trapeze (check the image).

I have to express the perimeter using only 1 variable (i was thinking c) and then derivate the perimeter and equal it to 0 to get the extremes..

how to express the perimeter using only c?

2. ok I tried something..

$b^2 = x \cdot 28$

then the perimeter is equal to:

$o = a + b + c + d = 2r + 2b + c$

$c = 28 - 2x$
$b = \sqrt{28x}$

therefore:

$o(x) = 28 + 2\sqrt{28x} + 28 - 2x = 56 - 2x + 2\sqrt{28x}$

$o'(x) = \frac{1}{2}\cdot 2(28x)^{-\frac{1}{2}} \cdot 28 - 2 = \frac{28}{\sqrt{28x}} -2$

to find the extremes:

$o'(x) = 0$
$\frac{28}{\sqrt{28x}} -2 = 0$

$28 = 2\sqrt{28x} /^2$
$784 = 112x$
$x = 7$

$o(x) = 56 - 2x + 2\sqrt{28x}$

$o(7) = 56 - 2 \cdot 7 + 2\sqrt{28 \cdot 7} = 42 + 28 = 70$

To check wether the extreme is max or min:

$o''(x) = (\frac{28}{\sqrt{28x}} -2)' = 28 \cdot (-\frac{1}{2})\cdot (28x)^{-\frac{3}{2}} \cdot 28 = -\frac {392}{\sqrt{(28x)^3}} (< 0) \Rightarrow Maximum$

can someone check if I'm right please :P