# Optimization question on graph

I was having trouble with this question.

A triangle has one side parallel to the x-axis, two vertices on the part of the parabola

$$y =3 − {x^2\over 12}$$ above the x-axis and the third vertex at the origin. Find two vertices so that the triangle has largest area.

I concluded that one of the endpoints must either be on $x = 6$ or $x = -6$ because one side must be parallel to the x-axis but I don't know where to go from there. Thank you.

The triangle is of height $y$ and width $2x$. Its area is $xy$. Do you see why? Hint: $y(x)$ is a symmetric parabola with respect to the $y$ axis. Therefore if a vertex is at $(x_0,y_0)$, and if the side is parallel to the $x$-axis then $(-x_0,y_0)$ has to be the other vertex. Now it's easy to see that the width of the triangle is twice the distance from $0$ to $x_0$, and its height is $y$.

Then the area $A(x)=3x-\dfrac{x^3}{12}$. You can now maximize it.

Clearly the point $C$ has coordinates $(x_0,y_0)$. $B$ is the other vertex. If $\overline{BC}\parallel x$-axis then $B$ has $y$-coordinate $y_0$. Furthermore, taking into account that the parabola is symmetric with respect to the $y$-axis, we can say that $y(x)=y(-x)$, therefore it follows that the $x$-coordinate of $B$ must be $-x_0$.
Now the area of a triangle is $\dfrac{\text{base}\times\text{height}}{2}$. The base is $BC=2x$, the height is obviously $y$.
• @user3495234 it's just $\dfrac{b·h}{2}$, right? The height is the distance $h$ from $A$ to $\overline{BC}$, which is simply $y_0$ given a $x_0$. The base is the distance $b$ from $B$ to $C$, which is $x_0-(-x_0)=2x_0$. Now area is just $xy$, but you already have $y$ so the area depends just on $x$. – Vladimir Vargas Nov 25 '14 at 2:37