Question

question
find the area of the rectangle of largest area that has its base on the ( x )-axis and its other 2 vertices above the ( x )-axis and lying on the parabola ( y = -2x^2 + 486 ).
answer attempt 1 out of 2

Explanation:

Step1: Define rectangle dimensions

Let the right vertex on the parabola be $(x, y)$. Since the parabola is symmetric about the y-axis, the base of the rectangle is $2x$, and height is $y=-2x^2+486$. The area function is:
$A(x) = 2x \cdot (-2x^2 + 486) = -4x^3 + 972x$

Step2: Find critical points

Take derivative of $A(x)$ and set to 0:
$A'(x) = -12x^2 + 972$
$-12x^2 + 972 = 0$
$x^2 = \frac{972}{12} = 81$
$x = 9$ (we take positive $x$ as length is positive)

Step3: Verify maximum area

Second derivative test: $A''(x) = -24x$. At $x=9$, $A''(9) = -216 < 0$, so this is a maximum.

Step4: Calculate maximum area

Substitute $x=9$ into $A(x)$:
$A(9) = -4(9)^3 + 972(9) = -4(729) + 8748 = -2916 + 8748$

Answer:

$5832$