Find the area of the largest rectangle with one corner at the origin,

Question

Find the area of the largest rectangle with one corner at the origin, the opposite corner in the first quadrant on the graph of the parabola f(x)= 162-6x^2  and sides parallel to the axes. 

The maximum possible area is 

Answer 1

Answer 2

With the upper right corner of the rectangle touching the parabola f(x)= 162-6x², the base can be expressed as x, and the height as 162-6x². The area would be the product of these two quantities, which can be expressed as A(x) = x(162-6x²).

This simplifies to A(x) = 162x - x³.

We then need to find the derivative of A(x) and find out where it is level [A'(x) = 0].

Using the power rule,

A'(x) = 162 - 3x²

Set it equal to 0

0 = 162 - 3x²

3x² = 162

x² = 54

x = ±3√6 ≈ ±7.3485.

The negative value can be eliminated because we are restricted to the first quadrant. To make sure we've done our due diligence, we need to make sure that this is a maximum. To do that, we find the second derivative. If it is concave down at that point [A''(x) < 0], then it is a maximum.

A''(x) = -6x²

A''(x) = -6(3√6)² = -324.

So we've verified that this is indeed a maximum. Now all that remains is to put that x value into A(x).

A(3√6) = 162(3√6) - (3√6)³ ≈ 793.635