A housing society wants to commission a swimming pool

Case study 2:

A housing society wants to commission a swimming pool for its residents. For this, they have to purchase a square piece of land and dig this to such a depth that its capacity is 250 cubic metres. Cost of land is Rs 500 per square metre. The cost of digging increases with the depth and cost for the whole pool is Rs 4000 (depth)²

Suppose the side of the square plot is x metres and depth is h metres. On the basis of the above information, answer the following question:

(i) Write cost C(h) as function in terms of h.

(ii) Find critical point.

(iii) (a) Use second derivative test to find the value of h for which cost of constructing the pool is minimum. What is the minimum cost of construction of the pool ?

(iii) (b) Use first derivative test to find the depth of the pool so that cost of construction is minimum. Also, find relation between x and h for minimum cost. [CBSE 2023]

Solution:

Given, side of square plot = x m

Depth of the pool = h m

Volume of pool = l×b×h

⇒ x × x × h = 250

⇒ x² h = 250

⇒ x² = 250/h

Digging cost = 4000 (depth)² = 4000(h)²

Total cost of pool with plot price and digging cost

C = 500 x² + 4000(h)²

⇒ C(h) = 500 × 250/h + 4000 h²

⇒ C(h) = 125000/h + 4000 h²

(ii) From (i)

C(h) = 125000/h + 4000 h²

Differentiating with respect to h

$\dfrac{d}{dh}C = - \dfrac{125000}{h^2} + 8000 h$ . . . . (i)

For max and minimum $\dfrac{d}{dh}C = 0$

$- \dfrac{125000}{h^2} + 8000 h = 0$

$\Rightarrow 8000 h = \dfrac{125000}{h^2}$

$\Rightarrow h^3 = \dfrac{125000}{8000}$

$\Rightarrow h^3 = \dfrac{125}{8}$

$\Rightarrow h = \dfrac{5}{2} = 2.5$

Hence the critical point is = 5/2 = 2.5

(iii) Taking equqtion (i)

$\dfrac{d}{dh}C = - \dfrac{125000}{h^2} + 8000 h$

Again differentiate with respect to x

$\dfrac{d^2}{dx^2}C = \dfrac{250000}{h^3} + 8000$

at h = 5/2 = 2.5

$\dfrac{d^2}{dx^2} C = \dfrac{250000}{(2.5)^3} + 8000$

$\Rightarrow \dfrac{d^2}{dx^2} C = \dfrac{250000}{2.5\times 2.5\times 2.5} + 8000$

$\Rightarrow \dfrac{d^2}{dx^2} C = 16000+8000 = 24000>0$

Hence the Cost of pool is minimum when h = 5/2 = 2.5

Minimum cost C(h) = 125000/h + 4000 h²

putting h = 2.5

C(h) = 125000/2.5 + 4000 (2.5)²

⇒ C(h) = 50,000 +25,000

Minimum cost C(h) = Rs 75,000

(iii) (b) C = 500 x² + 4000(h)²

⇒ C(h) = 500 × 250/h + 4000 h²

⇒ C(h) = 125000/h + 4000 h²

Differentiating with respect to h

$\dfrac{d}{dh}C = - \dfrac{125000}{h^2} + 8000 h$ . . . . (i)

For max and minimum $\dfrac{d}{dh}C = 0$

$- \dfrac{125000}{h^2} + 8000 h = 0$

$Rightarrow 8000 h = \dfrac{125000}{h^2}$

$\Rightarrow h^3 = \dfrac{125000}{8000}$

$\Rightarrow h^3 = \dfrac{125}{8}$

$\Rightarrow h = \dfrac{5}{2} = 2.5$

(a) Close to h = 2.5 and left to 2.5, h = 2

$\dfrac{d}{dh}C = - \dfrac{125000}{(2)^2} + 8000 \times 2$

$\Rightarrow \dfrac{d}{dh}C = - 31,250+ 16,000 = -15,250<0$

(b) Close to h = 2.5 and right to 2.5, h = 3

$\dfrac{d}{dh}C = - \dfrac{125000}{(3)^2} + 8000 \times 3$

$\Rightarrow \dfrac{d}{dh}C = - 13,888.89+ 24,000 = 10,111.11>0$

Note that for values close to 2.5 and to the right of 2.5, f ′(x) > 0 and for values close
to 2.5 and to the left of 2.5, f ′(x) < 0. Therefore, by first derivative test, x = 2.5 is a point
of local minima and local minimum value is

f(2.5) = Rs 75,000

The relation between x and h for minimum cost

C = 500 x² + 4000(h)²

⇒ 75,000 = 500 x² + 4000(h)²

Divide by 5000

⇒ 50 = x ² + 8(h)²

Hence the required relation is

x ² + 8(h)² = 50

Case study 1:

In a group activity class, there are 10 students whose age are 16, 17, 15, 14, 19, 17, 16, 19, 16 and 15 years. One student is selected at random such that each has equal chance of being chosen and age of the student is recorded.

On the basis of the above information, answer the following question:

(i) Find the probability that the age of the selected student is a composite number.

(ii) Let X be the age of the selected student. What can be the value of X ?

(iii) (a) Find the probability distributionof random variable X and hence find the mean age.

(b) A student was selected at random and his age was found to be greater than 15 years. Find the probability that his age is a prime number. [CBSE 2023]

Solution: For solution click here

Case study 3:

In an agricultural institute, scientists do experiments with varieties of seeds to grow them in different environments to produce healthy plants and get more yield.

A scientist observed that a particular seed grew very fast after germination. He had recorded growth of plant since germinantion and he said that its growth can defined by the function

$f(x) = \frac{1}{3} x^3 - 4x^2 + 15 x + 2, 0\leq x \leq 10$