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The figure shows a right triangle in the first quadrant. One side of the triangle is along the x-axis and the hypotenuse runs from the origin to a point on the parabola $y= 4−x^2$. Find the $x$ and $y$ coordinates that maximize the area of the triangle.

It's been a while since I did optimization, and am unsure as to how to go about doing this.

nmasanta
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caseyschlenker
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  • Write the area of the triangle in terms of $x$ and $y$. Then substitute $y=4-x^2$ for $y$ to get an expression in terms of $x$. Use calculus to find the maximum of this expression. – angryavian Apr 10 '21 at 17:30

2 Answers2

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The area of our trigangle $A = \frac 12 xy = \frac {1}{2} x(4-x^2)$

To find the extreme values of $A,$ take the derivative and set it equal to 0.

$A' = \frac 12 (4-3x^2) = 0\\ x = \sqrt {\frac 43}$

Now, if we are going to be good mathematicians we should verify whether this is a maximum or a minimum, but I will leave this as a exercise for the reader.

user317176
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The area of the triangle is

$$ S = \frac{1}{2} x y = \frac{1}{2} x(4 - x^2) = 2x - \frac{1}{2} x^3.$$

Hence, the maximum of $S$ occurs at $$ x = \pm \frac{2}{\sqrt{3}}$$

and the coordinates of $x$ and $y$ are

$$ (x,y) = \left ( \pm \frac{2}{\sqrt{3}} , \frac{8}{3} \right ).$$

Simon Pun
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