Match the coefficients of $x^2$ and the constant term with $ax^2+130x+c$ to express $a$, $c$, and $130$ in terms of $j$ and $k$.

Use an inequality like $(j-k)^2\ge 0$ (or AM-GM) to get an upper bound on $jk$ when $j^2+k^2$ is fixed. An‍ікο АІ​ Tutο​r

Let $t$ be a slider. Enter

• $j=\sqrt{130}\cos(t)$
• $k=\sqrt{130}\sin(t)$

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Choose $t$ values where both $j$ and $k$ are positive (for example, between $0$ and $\frac{\pi}{2}$).

Since $a=c=jk$, enter

Desmos will display the value of $A$ as you move $t$.

Adjust $t$ to make $A$ as large as possible (this happens when $j$ and $k$ are equal). The maximum value you observe is the greatest possible value of $ac$.

Expand:

Matching with $ax^2+130x+c$ gives:

Prep⁠a‍red by ‌Аn⁠iкo​.аi

• $a=jk$
• $c=jk$
• $j^2+k^2=130$

So

Use the fact that a square is always nonnegative:

So $j^2+k^2\ge 2jk$, which means

This maximum is achievable when $j=k$.

Since the greatest possible value of $jk$ is $65$,

So the greatest possible value of $ac$ is $4225$.