Question

1. find the maximum area of a right triangle with hypotenuse 1 m. (make sure to justify why your answer corresponds to an absolute maximum.)

Explanation:

Step1: Let the legs of the right - triangle

Let the legs of the right - triangle be \(x\) and \(y\). By the Pythagorean theorem, \(x^{2}+y^{2}=1\), so \(y = \sqrt{1 - x^{2}}\). The area of the right - triangle \(A=\frac{1}{2}xy\). Substitute \(y=\sqrt{1 - x^{2}}\) into the area formula, we get \(A(x)=\frac{1}{2}x\sqrt{1 - x^{2}}\), where \(0\lt x\lt1\).

Step2: Differentiate the area function

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = \frac{1}{2}x\) and \(v=\sqrt{1 - x^{2}}=(1 - x^{2})^{\frac{1}{2}}\).

\(u^\prime=\frac{1}{2}\), \(v^\prime=\frac{1}{2}(1 - x^{2})^{-\frac{1}{2}}\times(- 2x)=-\frac{x}{\sqrt{1 - x^{2}}}\)

\(A^\prime(x)=\frac{1}{2}\sqrt{1 - x^{2}}+\frac{1}{2}x\times(-\frac{x}{\sqrt{1 - x^{2}}})=\frac{1 - 2x^{2}}{2\sqrt{1 - x^{2}}}\)

Step3: Find the critical points

Set \(A^\prime(x) = 0\), then \(1-2x^{2}=0\) (since the denominator \(2\sqrt{1 - x^{2}}\gt0\) for \(0\lt x\lt1\)). Solving \(1 - 2x^{2}=0\) gives \(x=\frac{\sqrt{2}}{2}\) (we ignore \(x =-\frac{\sqrt{2}}{2}\) because \(x\gt0\)).

Step4: Use the second - derivative test

First, find the second - derivative.

Rewrite \(A^\prime(x)=\frac{1 - 2x^{2}}{2}(1 - x^{2})^{-\frac{1}{2}}\)

Using the product rule again: let \(u=\frac{1 - 2x^{2}}{2}\), \(u^\prime=-2x\); \(v=(1 - x^{2})^{-\frac{1}{2}}\), \(v^\prime=x(1 - x^{2})^{-\frac{3}{2}}\)

\(A^{\prime\prime}(x)=-x(1 - x^{2})^{-\frac{1}{2}}+\frac{1 - 2x^{2}}{2}\times x(1 - x^{2})^{-\frac{3}{2}}=\frac{-x(1 - x^{2})+\frac{1}{2}x(1 - 2x^{2})}{(1 - x^{2})^{\frac{3}{2}}}=\frac{-x + x^{3}+\frac{1}{2}x - x^{3}}{(1 - x^{2})^{\frac{3}{2}}}=\frac{-\frac{1}{2}x}{(1 - x^{2})^{\frac{3}{2}}}\)

When \(x = \frac{\sqrt{2}}{2}\), \(A^{\prime\prime}(\frac{\sqrt{2}}{2})=\frac{-\frac{1}{2}\times\frac{\sqrt{2}}{2}}{(1-\frac{1}{2})^{\frac{3}{2}}}\lt0\). So \(x=\frac{\sqrt{2}}{2}\) is a local maximum.

Since the function \(A(x)\) is continuous on the open interval \((0,1)\) and there is only one critical point, this local maximum is an absolute maximum.

When \(x=\frac{\sqrt{2}}{2}\), \(y=\sqrt{1 - (\frac{\sqrt{2}}{2})^{2}}=\frac{\sqrt{2}}{2}\)

The area \(A=\frac{1}{2}\times\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2}=\frac{1}{4}\)

Answer:

The maximum area of the right - triangle is \(\frac{1}{4}\text{ m}^2\)