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The hypotenuse of a right angle triangle measures 12 cm. What size angles would produce the maximum perimeter?

I got to point where I take the derivative and get $12(\cos\theta-\sin\theta)=0$, not sure where to go from there.

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The minimum perimeters occur when one of the sides is 12 cm and the other is infinitesimally short. Any deviation from this adds to the perimiter. Since the situation is symmetric, the maximum must occur when the angle is isosceles: a 12 cm hypotenuse and equal sides, which are $\frac{1}{\sqrt 2}\times 12$ long. – Kaz May 2 '13 at 5:48
Note also that if we fix the length of the hypotenuse, then all the possible right triangles have right corners which trace out a circle. The right angle corner of every possible right triangle is 6cm from the center of the hypotenuse, that being the radius. – Kaz May 2 '13 at 5:54
up vote 13 down vote accepted

We want to maximize $12(\sin\theta+\cos\theta)$. So we want to maximize $\sin\theta+\cos\theta$.

Equivalently, we want to maximize $(\sin \theta+\cos\theta)^2$. So we want to maximize $\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta$.

So we want to maximize $\sin 2\theta$. This happens when $2\theta=\frac{\pi}{2}$.

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Thank you for the response. – adeshina lawel May 2 '13 at 4:49
Andre you always amaze me. How do you think out of nowhere of such creative solutions? – Ovi May 2 '13 at 4:51
You are welcome. You are presumably intended to use the calculus. Note that to be sure you really have the max at $\pi/4$, you can use the fact that $\cos x-\sin x$ is positive up to $\pi/4$, then negative, so the function is incrasing, then decreasing. – André Nicolas May 2 '13 at 4:53
@Ovi: It is not out of nowhere, I think of this one as pretty standard. Every so often, there does come an interesting idea. But regrettably not often enough! – André Nicolas May 2 '13 at 4:56
Well I better keep this one in mind! I've never seen this technique before. And good luck on coming up with interesting ideas! – Ovi May 2 '13 at 5:00

Perimeter of a triangle $$ P = a+b+12 $$ which needs to be maximized with constraint $$ a^2+b^2=144 $$ Easiest way to do that is constructing Lagrangian using multipliers $$ L = a+b+12-\lambda\left( a^2+b^2-144\right) $$ so $$ \frac {\partial L}{\partial a} = 1-2\lambda a=0 \\ \frac {\partial L}{\partial b} = 1-2\lambda b=0 \\ \frac {\partial L}{\partial \lambda} = a^2+b^2-144=0 $$ One can easily find that $$ a = b = \frac {12}{\sqrt 2} $$ And therefore $\theta = \frac \pi 4$, since triangle is isosceles.


One can also see that answer doesn't really depend on the value of hypotenuse, and problem can be formulated as "Find angles of the right angle triangle with given hypotenuse $c$ and maximum possible perimeter." All of those triangles will be isosceles.

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Thanks a lot, this and the bottom response helped me. – adeshina lawel May 2 '13 at 4:41
+1 for the note. The maximum perimeter will always come from a right triangle. – Erick Robertson May 2 '13 at 12:25

Now you need to solve $\cos\theta-\sin\theta = 0$, i.e. $\cos\theta=\sin\theta$, so $$\tan\theta=1$$

From here, since $0 < \theta < \frac{\pi}{2}$, $$\theta=\frac{\pi}{2}-\theta=\frac{\pi}{4}$$

To check that this is indeed the maximum, the second derivative test is needed.

$$\left. \frac{d}{d\theta}[12(\cos\theta-\sin\theta)]\right|_{\theta=\frac{\pi}{4}} = \left. -12(\sin\theta + \cos\theta)\right|_{\theta=\frac{\pi}{4}} = -12\sqrt{2}<0$$ Hence $\theta=\frac{\pi}{4}$ is the point of maximum.

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Let's call the legs $x$ and $y$, and the perimeter $P$. So $$P=x+y+12$$ Draw the triangle, and by the Pythagorean theorem you should be able to see that $$ y=\sqrt{144-x^2} $$ Now substitute back to get $$P=x+\sqrt{144-x^2}+12 $$ Therefore, $$P'=1-\frac x{\sqrt{144-x^2}}$$ Now set this equal to $0$ and you should get $x$ to be $6\sqrt 2$. Now you have the hypotenuse and the adjacent leg, so $$\cos \theta=\frac {6\sqrt 2}{12}$$ or $$\cos \theta = \frac {\sqrt 2}2 $$ Therefore, $\theta=45^\circ$ or $\frac \pi 4$

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+1 for an example that uses only maths that a year 11 student would understand. (I'm being serious here. I admire simple solutions.) – daviewales May 2 '13 at 10:53
Thanks, I am in year 11 myself. – Ovi May 2 '13 at 23:35
Haha. I suppose that explains it. It's still very cool. – daviewales May 4 '13 at 1:07
Haha yea thanks – Ovi May 4 '13 at 2:17

So, the perimeter will be $12(1+\cos\theta+\sin\theta)=12+12(\cos\theta+\sin\theta)$

Now, $\sin\theta+\cos\theta=\sqrt2\cos(\theta-\frac\pi4)$ which will be maximum if $\theta-\frac\pi4=2n\pi$ where $n$ is any integer

So, $\theta=2n\pi+\frac\pi4$

As $0<\theta<\frac\pi2, \theta=\frac\pi4$

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