# A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y = 12 − x^2. What are the dimensions of such a rectangle with the greatest possible area?

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Steve M Share
Dec 21, 2016

The greatest area occurs when the rectangle has a width of 4 and a height of 8 leading to a maximum area of 32

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}P \left(x y\right) & \text{coordinate of the right hand corner" \\ A & "Area of Rectangle}\end{matrix}\right.$

$P$ lies on the parabola and $y = 12 - {x}^{2}$, so $P = P \left(x , 12 - {x}^{2}\right)$

Due to symmetry The width of the rectangle is half the distance between P and the y-axis, ie

Width = $2 x$ and Height=$y$

Hence the Area of the rectangle is:

$\setminus \setminus \setminus \setminus \setminus A = W \mathrm{di} t h \times H e i g h t$
$\therefore A = 2 x y$
$\therefore A = 2 x \left(12 - {x}^{2}\right)$
 :. A = 24x-2x^3)  ..... [1]

We are asked to maximise the Area as $x$ changes so hopefully we can identify a critical point of $A$ associated with a maximum, So we need to find $\frac{\mathrm{dA}}{\mathrm{dx}}$

Differentiating [1] wrt $x$
$\therefore \frac{\mathrm{dA}}{\mathrm{dx}} = 24 - 6 {x}^{2}$ ..... [2]

At a critical point, $\frac{\mathrm{dA}}{\mathrm{dx}} = 0$

$\therefore 24 - 6 {x}^{2} = 0$
$\therefore 6 {x}^{2} = 24$
$\therefore {x}^{2} = 4$
$\therefore x = \pm 2$

Obviously $x$ must be positive (otherwise we have an imaginary rectangle with negative area for a box that has collapsed in on itself)

$\therefore x = 2$

We need to check if this is a max or a min, so differentiate [2] wrt $x$ to get;

$\therefore \frac{{d}^{2} A}{\mathrm{dx}} ^ 2 = - 12 x$
$\therefore \frac{{d}^{2} A}{\mathrm{dx}} ^ 2 = - 12 x < 0 \text{ when } x = 2$, confirming a max

When $x = 2$ we have:

Width = $2 \cdot 2 = 4$
Height = $12 - 4 = 8$
Area = $32$

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