Chapter 4.3 Exercise 7

Question:

Find the maximal area of a right triangle with hypotenuse of length 1.

Answer:

So, the formula of the hypotenuse can be expressed as \({x}^{2} + {y}^{2} = 1\). When rearranged, this formula can be expressed as \(y = \sqrt{1 - {x}^{2}}\)

The formula for the area of a right triangle is \(A = \frac{1}{2}xy\). By plugging the first equation into the second, we get:

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

In order to determine the maximum optimization, we take the derivative:

\[{A}^{\prime} = \frac{d}{dx}\left[ \frac{1}{2}x\sqrt{1-{x}^{2}} \right]\]

Use the product rule.

\[{A}^{\prime} = \frac{1}{2} \left( \frac{d}{dx}\left[x\right]\sqrt{1 - {x}^{2}} + x\frac{d}{dx}\left[\sqrt{1 - {x}^{2}} \right] \right)\]

Now apply the derivation using the power rule

\[{A}^{\prime} = \frac{1}{2} \left( \sqrt{1 - {x}^{2}} + x\frac{1}{2}{\left(1 - {x}^{2}\right)}^{\frac{-1}{2}} \left(-2x \right) \right)\]

Simplify:

\[{A}^{\prime} = \frac{1}{2} \left( \sqrt{1 - {x}^{2}} - \frac{{x}^{2}}{\sqrt{1 - {x}^{2}}} \right)\]

Well, this is a pretty complicated equation so we can solve it graphically. Since we are only intereseted in postive values we don’t worry about looking for negative roots.

Here we see that y equals zero somewhere around where x is equal to 0.7. We can take a close look at the data frame directly.

##         x            y
## 704 0.703  0.008142692
## 705 0.704  0.006172909
## 706 0.705  0.004194824
## 707 0.706  0.002208379
## 708 0.707  0.000213514
## 709 0.708 -0.001789830
## 710 0.709 -0.003801713
## 711 0.710 -0.005822196
## 712 0.711 -0.007851341
## 713 0.712 -0.009889210

Here we see that the closest value is 0.707…. but when we look at the formula we can see that this number is actually more accurately expressed as \(\frac{1}{\sqrt{2}}\)

So, to wrap this up, the maximum area of a right triangle, with a hypotenuse of 1, is 0.25

I know…. we could have just put the work in and solved the equation, but graphically is so much prettier…

\[0 = \frac{1}{2} \left( \sqrt{1 - {x}^{2}} - \frac{{x}^{2}}{\sqrt{1 - {x}^{2}}} \right)\]

\[\sqrt{1 - {x}^{2}} = \frac{{x}^{2}}{\sqrt{1 - {x}^{2}}}\]

\[1 - {x}^{2} = {x}^{2}\] \[1 = 2{x}^{2}\]

\[x = \frac{1}{\sqrt{2}}\]