Case study:

Read the following passage and answer the questionS given below:

The relation between the height of the plant(‘y’ in cm) with respect to its exposure to the sunlight is governed by the following equation $y = 4x - \frac{1}{2}x^2$ , where ‘x’ is the number of days exposed to the sunlight, for x ≤ 3.

(i) Find the rate of growth of the plant with repect to the number of days exposed to the sunlight.

(ii) Does the rate of growth of the plant increase or decrease in the first three days ? what will be the height of the plant after 2 days ?

(iii) What is the number of days it will take for the plant to grow to the maximum height ?

(iv) What is the maximum height of the plant ?

Solution:

Given that, $y = 4x - \frac{1}{2}x^2$ , where ‘x’ is the number of days exposed to the sunlight, for x ≤ 3.

(i) The rate of growth of the plant with respect to the number of days exposed to sunlight is given by

$\frac{dy}{dx} = 4 - x$

(ii) Let rate of growth be represented by the function $g(x) = \frac{dy}{dx}$

Now, $g'(x) = \frac{d}{dx}(\frac{dy}{dx}) = -1\leq 0$

Hence g(x) decreases.

So the rate of growth of the plant decreases for the first three days.

Height of the plant after 2 days is $y = 4\times 2 - \frac{1}{2}(2)^2$

= 8 – 2 = 6 cm.

(iii) Since, $\frac{dy}{dx} = 4 - x$

For maxima and minima $\frac{dy}{dx}=0$

4 – x = 0

⇒ x = 4

Again differentiate with respect to x

$\frac{d^2y}{dx^2}= -1\leq 0$

Hence, height of the plant is maximum when x = 4 cm.

Plant get maximum height in 4 days

(iv) For maximum height of the plant putting x = 4 in $y = 4x - \frac{1}{2}x^2$

Now, $y = 4\times 4 - \frac{1}{2}(4)^2$

= 16 – 8 = 8 cm

Case study:

Dr. Ritam residing to Delhi went to see an apartment of 3 BHK in Noida. The window of the house was in the form of a rectangle surmounted by a semicircular opening having a perimeter of the window 10 m as shown in figure.