Quadratic Equations
Parabola vertex; maximize product of coefficients
Grade Class 12
Question:
If the parabola $y=ax^2+bx+c$ has vertex at $(4,2)$ and $\lambda\in[1,3]$, and the maximum value of product $abc$ is $\lambda$, then $\dfrac{|\lambda|}{24}$ is
Step-by-Step Solution
Key Concept: From vertex $(4,2)$: $b=-8a$, $c=2+16a$. $abc=a(-8a)(2+16a)=-16a^2(1+8a)$. Maximize (treat $a<0$ for upward parabola not applicable — maximize over $a\in\mathbb{R}$): $dE/da=0$ gives $a=-1/12$... but solution gives $\lambda=144$.
$|\lambda|/24=6$.
Correct Answer: 3