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My problem it says as follows: They give me a 1 meter ruler which has its endpoints on the sides of a right triangle. then they ask me what position of the ruler gives maximum area for the resulting triangle and also they ask me to find the maximum area. No clue about even how to start this . Cheers in advance |
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Let $a,b$ the sides of the right triangle. Then $a^2+b^2=1$, so $b=\sqrt{1-a^2}$, and so the area of the triangle is $\frac12a\sqrt{1-a^2}$. You'll want to find $0<a<1$ that maximizes this. (Hint: To maximize such an expression, you should examine its first derivative.) Here's a slightly different approach, using no calculus. Since $a^2+b^2=1$ and we need $0<a<1$ and $0<b<1$ to have a triangle in the first place, then we have $a=\sin\theta$ and $b=\cos\theta$ for some $0<\theta<\frac\pi2$. (Do you see why?) The area of the triangle then depends only on $\theta$, and is given by $\frac12ab=\frac12\sin\theta\cos\theta.$ Using double-angle formula for sine, the area of the triangle is $$\frac14\sin(2\theta).$$ Since $\sin\phi$ takes on its maximum value over $(0,\pi)$ precisely when $\phi=\frac\pi2$, it follows that the area of the triangle is maximized when $\theta=\frac\pi4$. (Why?) What is the maximum area, then? What is the position of the ruler that gives you that area--in particular, what can you say about $a$ and $b$ when $\theta=\frac\pi4$? |
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The tag says calculus, but we will use less machinery. Let the legs of the triangle be $a$ and $b$. We want to maximize $\frac{1}{2}ab$, given that $a^2+b^2=1$. Equivalently, we want to maximize $2ab$. Note that $$2ab=(a^2+b^2)-(a-b)^2=1-(a-b)^2.$$ So $2ab$ is $1$ minus a square. To maximize $2ab$, we need to make $(a-b)^2$ as small as possible. The smallest possible value of $(a-b)^2$ is $0$. Thus the maximum of $2ab$ is $1$, and therefore the maximum area is $\frac{1}{4}$. This is achived when $a=b$. |
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