Find the maximal area of a right triangle with hypotenuse of length 6.
I've labeled my triangle with Z being the hypotenuse and the two sides X and Y.
I know
$$A = BH/2 = XY/2$$
Using the Pythagorean theorem I find that
$$ Y = \sqrt{36-X^2}$$
I then get
$$A = \frac{X\sqrt{36-X^2}}{2}$$
I'm getting the derivative of
$$A'=\frac{-X^2 + 18} {\sqrt{36-X^2}}$$
After setting this equal to 0 to solve for X, i'm getting
$$X^2 = -18$$
Which would give an imaginary number and is incorrect. Where am I going wrong?
Thanks!
Best Answer
You are correct, the area is given by $$ A(x) = \frac{x\sqrt{36-x^2}}{2} $$ but derivative is from the product rule: $$ A'(x) = \frac{\sqrt{36-x^2}}{2} + \frac{x \times (-2x)}{2 \times 2\sqrt{36-x^2}} = \frac{18-x^2}{\sqrt{36-x^2}} $$ So $A'(x) = 0$ if and only if $18=x^2$ or $x = \pm \sqrt{18} = \pm 3\sqrt{2}$, of which you obviously only need the positive one.