# A rectangle is constructed with it's base on the x-axis and the two of its vertices on the parabola y=49 - x^2. What are the dimensions of the rectangle with the maximum area?

Jun 22, 2015

In other words, we're constructing a rectangle under a dome-shaped form.

#### Explanation:

The parabola is a 'mountain'-type (because the coefficient of ${x}^{2}$ is negative. Also, it is symmetrical in respect to the $y$-axis, because there is no $x$-term.
We can now simplify the problem to finding a rectangle with vertices at $\left(0 , 0\right) , \left(x , 0\right) , \left(0 , y\right) \mathmr{and} \left(x , y\right)$ and then double the $x$-values.
The area will then be $A = x \cdot y$
If we substitute the equation of the parabola for $y$:
$A = x \cdot \left(49 - {x}^{2}\right) = 49 x - {x}^{3}$

To find the extremes (max of min) we need the derivative and set it to zero:
$A ' = 49 - 3 {x}^{2} = 0 \to {x}^{2} = \frac{49}{3} \to x = \sqrt{\frac{49}{3}} \approx 4.04 \ldots$
(remember we will have to double that)

Use this in the original function:

$y = 49 - {x}^{2} = 49 - \frac{49}{3} = \frac{98}{3} \approx 32.67$

Dimensions will be $8.08 \text{ x } 32.67$