the price p per unit at which a company can sell all that it produces is given by the function p(x) = 300-4x. the costs function is c(x) = 500+28x where x is the number of units produced. find x so that the profit is maximum

profit = sales-cost

sales=quantity*price = x(300-4x)
cost = 500+28x

profit = (300x-4x^2)-(500+28x)
= -4x^2 + 272x - 500
profit is max when -8x+272=0, or
x = 34

CASE STUDY: 1

The bulbs manufactured by a company gave a mean life of 3000 hours with standard
deviation of 400 hours. If a bulb is selected at random, what is the probability it will
have a mean life less than 2000 hours?
Question:
1) Calculate the probability.
2) In what situation does one need probability theory?
3) Define the concept of sample space, sample points and events in context of
probability theory.
4) What is the difference between objective and subjective probability?
CASE STUDY : 2
The price P per unit at which a company can sell all that it produces is given by the
function P(x) = 300 — 4x. The cost function is c(x) = 500 + 28x where x is the number
of units produced. Find x so that the profit is maximum.
Question:
1) Find the value of x.
2) In using regression analysis for making predictions what are the assumptions
involved.
3) What is a simple linear regression model?
4) What is a scatter diagram method?
CASE STUDY : 3
Mr Sehwag invests Rs 2000 every year with a company, which pays interest at 10% p.a.
He allows his deposit to accumulate at C.I. Find the amount to the credit of the person
at the end of 5th year.
Question :
1) What is the Time Value of Money concept.
2) What do you mean by present value of money?
3) What is the Future Value of money.
4) What the amount to be credited at the end of 5th year.
CASE STUDY : 4
The cost of fuel in running of an engine is proportional to the square of the speed and is
Rs 48 per hour for speed of 16 kilometers per hour. Other expenses amount to Rs 300
per hour. What is the most economical speed?
Question:
1) What is most economical speed?
2) What is a chi-square test?
3) What is sampling and what are its uses.
4) Is there any alternative formula to find the value of Chi-square?

Define ownership cost? What are the various cash flow affecting ownership

cost?

Well, let's first start by finding the profit function. Profit is defined as the difference between revenue and costs. We know that revenue is the selling price multiplied by the number of units sold, which is given by p(x) * x. And costs are given by the function c(x). Therefore, the profit function is:

Profit(x) = (p(x) * x) - c(x)

Substituting the given functions for p(x) and c(x), we get:

Profit(x) = (300 - 4x) * x - (500 + 28x)

Expanding this equation, we get:

Profit(x) = 300x - 4x^2 - 500 - 28x

Simplifying, we have:

Profit(x) = -4x^2 + 272x - 500

To find the x-value that maximizes the profit, we can take the derivative of the profit function with respect to x and set it equal to zero:

Profit'(x) = -8x + 272 = 0

Solving for x, we have:

-8x = -272
x = -272 / -8
x = 34

So, the number of units produced that maximizes profit is 34. However, since you can't have a negative number of units, we can discard the negative solution.

Therefore, the number of units produced that maximizes the profit is x = 34.

To find the number of units x that maximizes the profit, we need to determine the profit function and then find the critical points.

The profit can be calculated by subtracting the cost function from the revenue function:
Profit = Revenue - Cost

The revenue function is given by p(x), which represents the price per unit multiplied by the quantity sold:
Revenue = p(x) * x = (300 - 4x) * x = 300x - 4x^2

The cost function is given by c(x) = 500 + 28x.

Now, let's calculate the profit function by subtracting the cost function from the revenue function:
Profit = Revenue - Cost = (300x - 4x^2) - (500 + 28x) = -4x^2 + 272x - 500

To find x that maximizes the profit, we need to find the critical points of the profit function. This can be done by taking the derivative of the profit function and finding where it equals zero.

Differentiating the profit function with respect to x:
d(Profit)/dx = -8x + 272

Setting the derivative equal to zero and solving for x:
-8x + 272 = 0
-8x = -272
x = 34

So, the critical point is x = 34. To verify that it is a maximum, we can check the concavity of the profit function. Taking the second derivative of the profit function:

d^2(Profit)/dx^2 = -8

Since the second derivative is a constant (-8), which is negative, it confirms that x = 34 is indeed a maximum point.

Therefore, to maximize the profit, the company should produce 34 units.