A triangle has one vertex at $(0,0)$, the other two on graph of $y=-2x^2+54$

Question

A triangle has one vertex at $(0,0)$ and the other two on graph of $y=-2x^2+54$ at $(x,y)$,$(-x,y)$ where $0<x<27$. The value of $x$ corresponding to maximum area?

I found out the Area as xy and took its derivative and equated it to zero from where I got the slope as -y/x. Since area is given to be maximum, I took the derivative again and tried solving using inequalities but unsuccessfully. Help would be appreciated !!

Please add the statement of the problem to the body of your post. — Another User, Aug 01 '23 at 18:13
Are you sure about $27$? I ask this because $\sqrt{27}$ looks much more natural. — José Carlos Santos, Aug 01 '23 at 18:21
Note that $y$ is a function of $x$. So your area $xy=x(-2x^2+54)$ — Andrei, Aug 01 '23 at 18:49
First see the comment of Andrei. Note that if, for a specific value for $~x,~$ the first derivative equals $~0~$ and the second derivative is negative, then you have a (local) maximum. — user2661923, Aug 01 '23 at 19:33
@Andrei yes, it was supposed to be root 27, there was an error in the question. could you tell me how you figured it out? — Mukesh Gopal, Aug 02 '23 at 02:19

score 0 · Answer 1 · answered Aug 01 '23 at 21:27

Let the vertices of the triangle be $(0,0)$, $(x,-2x^2+54)$, and $(-x,-2x^2+54)$ (using the fact that $y=-2x^2+54)$

Using the determinant method to find the area as a function of $x$;

$$\left|\frac{1}{2}\begin{pmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{pmatrix}\right|=f(x)$$

$$\left|\frac{1}{2}\begin{pmatrix}0 & 0 & 1\\x & -2x^2+54 & 1\\ -x & -2x^2+54 & 1\end{pmatrix}\right|=f(x)$$

$$f(x)= x(2x^2-54)$$

Now, to maximize this area, solve $f'(x)=0$;

$$f'(x)=\left[x(2x^2-54)\right]' = 6x^2 - 54 = 6(x^2 - 9) = 0$$

On solving, we get $x=\pm 3$. But as the condition is $x>0$, $x=3$ is your required value

A triangle has one vertex at $(0,0)$, the other two on graph of $y=-2x^2+54$

1 Answers1