Among all right triangles whose hypotenuse has a length of 12 cm, what is the largest possible perimeter?

Answer:Largest perimeter of the triangle =  [tex]P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)[/tex]Step-by-step explanation:We are given the following information in the question:Right triangles whose hypotenuse has a length of 12 cm.Let x and y be the other two sides of the triangle.Then, by Pythagoras theorem:[tex]x^2 + y^2 = (12)^2 = 144\\y^2 = 144-x^2\\y = \sqrt{144-x^2}[/tex]Perimeter of Triangle = Side 1 + Side 2 + Hypotenuse.[tex]P(x) = x + \sqrt{144-x^2} + 12[/tex]where P(x) is a function of the perimeter of the triangle.First, we differentiate P(x) with respect to x, to get,[tex]\frac{d(P(x))}{dx} = \frac{d(x + \sqrt{144-x^2} + 12)}{dx} = 1-\displaystyle\frac{x}{\sqrt{144-x^2}} [/tex]Equating the first derivative to zero, we get,[tex]\frac{dP(x))}{dx} = 0\\\\1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0[/tex]Solving, we get,[tex]1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0\\\\x = \sqrt{144-x^2}}\\\\x^2 = 144-x^2\\\\x = \sqrt{72} = 6\sqrt{2}[/tex]Again differentiation P(x), with respect to x, using the quotient rule of differentiation.[tex]\frac{d^2(P(x))}{dx^2} = \displaystyle\frac{-(144-x^2)^{\frac{3}{2}}-x^2}{(144-x)^{\frac{3}{2}}}[/tex]At x = [tex]6\sqrt{2}[/tex],[tex]\frac{d^2(V(x))}{dx^2} < 0[/tex]Then, by double derivative test, the maxima occurs at x = [tex]6\sqrt{2}[/tex]Thus, maxima occurs at x = [tex]6\sqrt{2}[/tex] for P(x).Thus, largest perimeter of the triangle =  [tex]P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)[/tex]