Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

Find the area of the largest rectangle with lower base on the

x

-axis and upper vertices on the curve

y = 12 - x^{2} .

Accepted Answer

Area of rectangle

$P$ lies on the parabola and

$y = 12 - x^{2}$ , so $P = P (x, 12 - x^{2})$

Due to symmetry, the width of the rectangle is half the distance between $P$ and the $y$ -axis, i.e.

width $= 2 x$ and height $= y$

Hence, the area of the rectangle is

$A = Width \times Height$

$∴ A = 2 x y$

$A = 2 x (12 - x^{2})$

$A = 24 x - 2 x^{3} . . . (i)$

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 $\frac{d A}{d x}$ .

Differentiating $(i)$ w.r.t. $x$

$∴ \frac{d A}{d x} = 24 - 6 x^{2} . . . (ii)$

At a critical point, $\frac{d A}{d x} = 0$

$∴ 24 - 6 x^{2} = 0$

$\Rightarrow 6 x^{2} = 24$

$\Rightarrow x^{2} = 4$

$∴ x = \pm 2$

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

$∴ x = 2$

We need to check if this is a maximum or a minimum, so differentiate $(ii)$ w.r.t. $x$ to get

$∴ \frac{d^{2} A}{d x^{2}} = - 12 x$

$\frac{d^{2} A}{d x^{2}} = - 12 x < 0$ when $x = 2$ ,

confirming a maximum

When $x = 2$ we have:

Width $= 2 \cdot 2 = 4$

Height $= 12 - 4 = 8$

Area $= 32$ sq. units

Find the area of the largest rectangle with lower base on the $x$ -axis and upper vertices on the curve $y = 12 - x^{2} .$

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y=12-x2.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Find the area of the largest rectangle with lower base on the $x$ -axis and upper vertices on the curve $y = 12 - x^{2} .$