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 two legs be \(x\) and \(y\)

By Pythagorean theorem, \(x^{2}+y^{2}=1\), and the area \(A=\frac{1}{2}xy\). From \(x^{2}+y^{2}=1\), we have \(y = \sqrt{1 - x^{2}}\). Then \(A(x)=\frac{1}{2}x\sqrt{1 - x^{2}},0<x<1\).

Step2: Differentiate \(A(x)\)

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

Step3: Find critical points

Set \(A^\prime(x) = 0\), then \(1-2x^{2}=0\) (since the denominator \(2\sqrt{1 - x^{2}}
eq0\) for \(0 < x<1\)). Solving \(1-2x^{2}=0\) gives \(x=\frac{\sqrt{2}}{2}\) (we ignore \(x =-\frac{\sqrt{2}}{2}\) as \(x\in(0,1)\))

Step4: Second - derivative test

Differentiate \(A^\prime(x)\) using the quotient - rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) where \(u = 1-2x^{2}\), \(u^\prime=-4x\), \(v = 2\sqrt{1 - x^{2}}\), \(v^\prime=\frac{-2x}{\sqrt{1 - x^{2}}}\)
\(A^{\prime\prime}(x)=\frac{-4x\times2\sqrt{1 - x^{2}}-(1 - 2x^{2})\times\frac{-2x}{\sqrt{1 - x^{2}}}}{4(1 - x^{2})}\)
Substitute \(x = \frac{\sqrt{2}}{2}\) into \(A^{\prime\prime}(x)\):
\(A^{\prime\prime}(\frac{\sqrt{2}}{2})<0\), so \(x=\frac{\sqrt{2}}{2}\) is a local maximum.
When \(x=\frac{\sqrt{2}}{2}\), \(y=\sqrt{1 - (\frac{\sqrt{2}}{2})^{2}}=\frac{\sqrt{2}}{2}\)
\(A=\frac{1}{2}\times\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2}=\frac{1}{4}\)

Also, we can use the AM - GM inequality. For non - negative numbers \(a\) and \(b\), \(\frac{a + b}{2}\geqslant\sqrt{ab}\), equality holds when \(a = b\).
Since \(x^{2}+y^{2}=1\), and \(A=\frac{1}{2}xy\). By \(x^{2}+y^{2}\geqslant2xy\) (equality when \(x = y\)), then \(xy\leqslant\frac{1}{2}\), so \(A=\frac{1}{2}xy\leqslant\frac{1}{4}\)

Answer:

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