How do you find the right triangle of maximum area if the sum of the lengths of the legs is 5?

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Answer 1 · 2015-06-07T04:37:51

Call x and y the 2 legs.

Area of the right triangle: $s = \frac{x . y}{2}$ $s = (x.y)/2$

Given: x + y = 5 --> y = 5 - x

Then, $s = \frac{x}{2} (5 - x) = - (\frac{1}{2}) (x^{2} - 5 x)$ $s = x/2(5 - x) = -(1/2)(x^2 - 5x)$

s maximum when the derivative s' = 0 -> 2x - 5 = 0 --> $x = \frac{5}{2}$ $x = 5/2$ .

The area is max when both legs are equals. $x = y = \frac{5}{2}$ $x = y = 5/2$

$s max = (1/2)(5/2)(5/2) = 25/8$

Check:

When x = 2, y = 3, the area is $s = (2(3))/2 =$ 3
When x = 1, and y = 4, the area is $s = (1/2)(4) = 2$