Question

the total profit 'y' in ₹ of a company is given by y= -x^2/400 +2x-80. the sale of x bottles is considered for the financial year 2011-12.find how many bottles the company must sell to achieve maximum profit​

Answer 1

Question :-

• The total profit 'y' in ₹ of a company is given by y= -x^2/400 +2x-80. The sale of x bottles is considered for the financial year 2011-12. Find how many bottles the company must sell to achieve maximum profit?

Answer :-

Given :

The total profit 'y' in ₹ of a company is given by

$\bf \:y = - \dfrac{ {x}^{2} }{400} + 2x - 80$

To Find :-

• How many bottles the company must sell to achieve maximum profit?

Definition :-

Local Maximum

• A function f(x) is said to have a local maximum at x=a if the value of f(a) is greater than all the values of f(x) in a small neighbourhood of x=a.
• Mathematically, f (a) > f (a – h) and f (a) > f (a + h) where h > 0, then a is called the point of local maximum.

Local Minimum

• A function f(x) is said to have a local minimum at x = a, if the value of the function at x = a is less than the value of the function at the neighboring points of x = a. Mathematically, f (a) < f (a – h) and f (a) < f (a + h) where h > 0, then a is called the point of local minimum.

A point of local maximum or a local minimum is also called a point of local extremum.

Working Rule to Determine the points of Local Maxima and Local Minima

1. First Derivative Test: 

If f'(a)  = 0 and f'(x)  changes it’s sign while passing through the point x = a, then

(i) f(x) would have a local maximum at x = a if f'(a–0)  > 0 and  f'(a+0) < 0. It means that f'(x) should change it’s sign from positive to negative e.g. f (x) = –x2 has local maxima at x = 0.

(ii) f(x) would have local minimum at x = a if f'(a–0) < 0  and f'(a+0) > 0 . It means that f'(x) should change it’s sign from negative to positive. e.g. f (x) = x2 has local minima at x = 0.

(iii) If f(x) doesn’t change it’s sign while passing through x = a, then f (x) would have neither a maximum nor minimum at x = a.

2. Second Derivative Test:

Step I: Let f(x) be a differentiable function on a given interval and let f’’ be continuous at stationary point. Find f ‘ (x) and solve the equation f ‘ (x) = 0 given let x = a, b, … be solutions.

Step II: Case (i) : If f ‘‘ (a) < 0 then f(a) is maximum.

            Case (ii): If f ‘’(a) > 0 then f(a) is minimum.

Note:

(i) If f’’(a) = 0 the second derivatives test fails in that case we have to go back to the first derivative test.

(ii) If f’’(a) = 0 and a is not a point of local maximum nor local minimum then a is a point of inflection.

Solution : -

$\bf \:y = - \dfrac{ {x}^{2} }{400} + 2x - 80$

Differentiate w. r. t. x

$\bf\implies \:\dfrac{dy}{dx} = - \dfrac{2x}{400} + 2.......(1)$

For maxima or minima,

$\bf \:put \: \dfrac{dy}{dx} = 0$

$\bf\implies \: - \dfrac{ 2x}{400} + 2 = 0$

$\bf\implies \:\dfrac{2x}{400} = 2$

$\bf\implies \:x = 400$

Now, again Differentiate (1) w. r. t. x, we get

$\bf\implies \:\dfrac{ {d}^{2} y}{d {x}^{2} } = - \dfrac{2}{400} < 0$

Hence, y is maximum when x = 400.

⇛ 400 bottles, the company must sell to achieve maximum profit.