How do you find the right triangle of maximum area if the sum of the lengths of the legs is 3?

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Answer 1 · 2015-06-03T18:54:11

We know that $x + y = 3$ $x+y=3$ and that the area $A = \frac{x \cdot y}{2}$ $A = (x*y)/2$ because of the right angle.

Let's use substitution and replace $y$ $y$ in the equation of the area :

$x + y = 3 \Leftrightarrow y = 3 - x$ $x+y=3<=>y=3-x$

$A = \frac{x \cdot (3 - x)}{2} = \frac{- x^{2} + 3 x}{2}$ $A = (x*(3-x))/2 = (-x^2+3x)/2$

The maximum area can be found by studying the sign of the derivative of the area :

$A'(x) = (-2x+3)/2$

The derivative $= 0$ when $-2x+3=0<=>x=3/2$ .
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Therefore, the area is maximal when $x=3/2$ :

$A=(-(3/2)^2+3*(3/2))/2 = (-(9/4)+(18/4))/2 = 9/8$ .