2. find the maximum area of a right triangle with hypotenuse 1 m. (make sure to justify why your answer…

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) into the area formula: (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). Let (u = \frac{1}{2}x) and (v=\sqrt{1 - x^{2}}=(1 - x^{2})^{\frac{1}{2}}). (u^\prime=\frac{1}{2}) and (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 - x^{2}-x^{2}}{2\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). Solving for (x), we get (x=\frac{1}{\sqrt{2}}) (since (x\gt0)). When (x = \frac{1}{\sqrt{2}}), (y=\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}=\frac{1}{\sqrt{2}})

Step4: Second - derivative test (or boundary behavior)

We can also use the second - derivative test. First, rewrite (A^\prime(x)=\frac{(1 - 2x^{2})(1 - x^{2})^{-\frac{1}{2}}}{2}) Using the quotient rule (\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}) where (u = 1 - 2x^{2}), (u^\prime=-4x), (v = 2(1 - x^{2})^{\frac{1}{2}}), (v^\prime=2\times\frac{1}{2}(1 - x^{2})^{-\frac{1}{2}}\times(-2x)) (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{1}{\sqrt{2}}) into (A^{\prime\prime}(x)): (A^{\prime\prime}\left(\frac{1}{\sqrt{2}}\right)\lt0), so (x = \frac{1}{\sqrt{2}}) is a local maximum. Another way: As (x\rightarrow0^{+}), (A(x)\rightarrow0); as (x\rightarrow1^{-}), (A(x)\rightarrow0).

Answer:

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