Question

if the length of the hypotenuse of a right triangle is 10, find the lengths of the other sides when the area is a minimum

Answer 1

A right triangle has one side equal to x, its hypotenuse equals 10. Then its third side is equal to $\sqrt{100-x^2}$ .

Area of this triangle is $A(x)=\frac{x\sqrt{100-x^2}}{2}$ .

Let is find minimum of the function $A(x)$ on interval (0;10) (the length of the side is a positive number and to can’t be greater than length of hypotenuse)

$A^\prime (x)=\left( \frac{x\sqrt{100-x^2}}{2} \right)^\prime = \frac{\sqrt{100-x^2}}{2}-\frac{x^2}{2\sqrt{100-x^2}}=\frac{100-2x^2}{2\sqrt{100-x^2}}$

We put $A^\prime (x)=0$

$100-2x^2=0$

$x^2=50$

$x=5\sqrt{2}$

We compute $A(x)$ at the critical point: $A(5\sqrt{2})=25$

Now we find $A(x)$ at the boundaries of the interval: $A(0)=A(10)=0$ .

Therefore, $x=5\sqrt{2}$ is the point of maximum and $A(x)$ is minimum at the boundaries, but when $x=0$ or $x=10$ there is no triangle.

So, there is no minimum area of this triangle.