Math Problem Statement

A company produces certain items. The manager maintains a record of cost and revenue of these items on a daily basis. The cost and revenue functions of the product are given by C(x) = 15x + 3000 and R(x) = 45x + 1500 respectively, where x is the number of items produced and sold. Answer the following: i) How many items must be produced so that the cost is less than 4500? ii) What will be the minimum quantities of items to be sold so that the revenue is greater than or equal to 10500 rupees? iii) If the number of items sold lies between 300 to 400, what will be the revenue? OR) How many items must be sold to gain some profit?

Solution

Let's analyze and solve the questions in the image you provided.

Given functions:

  • Cost function: C(x)=15x+3000C(x) = 15x + 3000
  • Revenue function: R(x)=45x+1500R(x) = 45x + 1500 where xx represents the number of items produced and sold.

Now let's solve each part:

i) How many items must be produced so that the cost is less than 4500?

We need to find xx such that the cost C(x)<4500C(x) < 4500.

C(x)=15x+3000C(x) = 15x + 3000 Setting the inequality: 15x+3000<450015x + 3000 < 4500 Solving for xx: 15x<4500300015x < 4500 - 3000 15x<150015x < 1500 x<150015=100x < \frac{1500}{15} = 100 Thus, the number of items must be less than 100 for the cost to be less than 4500.

ii) What will be the minimum quantities of items to be sold so that the revenue is greater than or equal to 10,500 rupees?

We need to solve for xx such that R(x)10500R(x) \geq 10500.

R(x)=45x+1500R(x) = 45x + 1500 Setting the inequality: 45x+15001050045x + 1500 \geq 10500 Solving for xx: 45x10500150045x \geq 10500 - 1500 45x900045x \geq 9000 x900045=200x \geq \frac{9000}{45} = 200 Thus, the minimum number of items to be sold for revenue to be greater than or equal to 10500 is 200 items.

iii) If the number of items sold lies between 300 to 400, then what will be the revenue?

We are asked to find the revenue when xx is between 300 and 400, i.e., for 300x400300 \leq x \leq 400.

Using the revenue function R(x)=45x+1500R(x) = 45x + 1500:

For x=300x = 300: R(300)=45(300)+1500=13500+1500=15000R(300) = 45(300) + 1500 = 13500 + 1500 = 15000

For x=400x = 400: R(400)=45(400)+1500=18000+1500=19500R(400) = 45(400) + 1500 = 18000 + 1500 = 19500

So, the revenue will be between 15000 and 19500 when the number of items sold lies between 300 and 400.

OR) How many items must be sold to gain some profit?

To make a profit, the revenue must exceed the cost, i.e., R(x)>C(x)R(x) > C(x).

Setting the inequality: 45x+1500>15x+300045x + 1500 > 15x + 3000 Solving for xx: 45x15x>3000150045x - 15x > 3000 - 1500 30x>150030x > 1500 x>150030=50x > \frac{1500}{30} = 50 Thus, the company must sell more than 50 items to gain some profit.


Would you like more details on any part of this? Here are some related questions for deeper understanding:

  1. How would the results change if the cost function had a different fixed cost?
  2. What would happen to the minimum quantity required for profit if the selling price per item was lowered?
  3. How would you compute the break-even point for this cost and revenue system?
  4. What is the profit if 400 items are sold?
  5. How do changes in fixed cost or variable cost affect the profit margin?

Tip: When solving inequalities, always check if the solution needs to be rounded or exact, especially in real-world scenarios like production or sales.

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Math Problem Analysis

Mathematical Concepts

Linear Functions
Inequalities
Revenue and Cost Analysis

Formulas

C(x) = 15x + 3000
R(x) = 45x + 1500
Profit condition: R(x) > C(x)

Theorems

Solving inequalities
Basic profit condition

Suitable Grade Level

Grades 9-12