Samacheer Kalvi Class 11 Business Maths Solutions Chapter 6 Applications of Differentiation Exercise 6.2

Question 1. The average cost function associated with producing and marketing x units of an item is given by \( AC = 2x - 11 + \frac{50}{x} \). Find the range of values of the output x, for which AC is increasing.
Answer: The average cost (AC) function is given by \( AC = 2x - 11 + \frac{50}{x} \).
For AC to be increasing, its derivative with respect to x must be greater than 0.
First, we find the derivative of AC:
\( \frac{d AC}{d x} = \frac{d}{dx}(2x - 11 + 50x^{-1}) \)
\( \frac{d AC}{d x} = 2 - 0 + 50(-1x^{-2}) \)
\( \frac{d AC}{d x} = 2 - \frac{50}{x^{2}} \)
Next, we set the derivative to be greater than zero for AC to increase:
\( 2 - \frac{50}{x^{2}} > 0 \)
\( 2 > \frac{50}{x^{2}} \)
Now, multiply both sides by \( x^{2} \) (since \( x^{2} \) is always positive, the inequality sign does not flip):
\( 2x^{2} > 50 \)
Divide by 2:
\( x^{2} > 25 \)
Take the square root of both sides:
\( x > \pm 5 \)
Since x represents the number of units produced, it must be a positive value. Therefore, we take the positive root.
So, \( x > 5 \). The average cost is increasing when the output x is greater than 5 units.
In simple words: We find out how the average cost changes as we make more items. If this change is positive, it means the cost is going up. We calculate this point and find that when we make more than 5 items, the average cost starts to increase.

🎯 Exam Tip: Remember that output (x) must always be a positive value in real-world scenarios, so always consider the positive root when solving for x in such problems.

Question 2. A television manufacturer finds that the total cost for the production and marketing of x number of television sets is \( C(x) = 300x^{2} + 4200x + 13500 \). If each product is sold for Rs. 8,400. show that the profit of the company is increasing.
Answer: The total cost function for producing x television sets is given by \( C(x) = 300x^{2} + 4200x + 13500 \).
The selling price of one product is Rs. 8,400.
So, the selling price for x products is \( 8400x \).
The profit function, P(x), is calculated as: Profit = Selling price - Total cost.
\( P(x) = 8400x - (300x^{2} + 4200x + 13500) \)
\( P(x) = 8400x - 300x^{2} - 4200x - 13500 \)
\( P(x) = -300x^{2} + 4200x - 13500 \)
To find out if the profit is increasing, we need to find the first derivative of the profit function, \( P'(x) \), and see if it is positive.
\( P'(x) = \frac{d}{dx}(-300x^{2} + 4200x - 13500) \)
\( P'(x) = -600x + 4200 \)
To find critical points, we set \( P'(x) = 0 \):
\( -600x + 4200 = 0 \)
\( 4200 = 600x \)
\( x = \frac{4200}{600} \)
\( x = 7 \)
This critical point \( x = 7 \) divides the possible output range (which must be \( x > 0 \)) into two intervals: \( (0, 7) \) and \( (7, \infty) \).
We need to test an x-value in the interval \( (0, 7) \) to see if the profit is increasing. Let's pick \( x = 2 \).
Substitute \( x = 2 \) into \( P'(x) \):
\( P'(2) = -600(2) + 4200 \)
\( P'(2) = -1200 + 4200 \)
\( P'(2) = 3000 \)
Since \( P'(2) = 3000 \) is a positive value, it means the profit is increasing when \( x \) is in the interval \( (0, 7) \). This confirms the profit is increasing up to 7 units, showing a positive trend in that range.
In simple words: We first write down the formula for how much profit the company makes. Then, we find out how this profit changes when they sell more TVs. We check a point where profit increases, and since the calculation shows a positive number, it means they are making more money for each additional TV sold up to a certain point.

🎯 Exam Tip: To show that profit is increasing, always calculate the first derivative of the profit function and demonstrate that it is positive for relevant values of the output (x).

Question 3. A monopolist has a demand curve \( x = 106 - 2p \) and average cost curve \( AC = 5 + \frac{x}{50} \), where p is the price per unit output and x is the number of units of output. If the total revenue is \( R = px \), determine the most profitable output and the maximum profit.
Answer: We are given the demand curve as \( x = 106 - 2p \).
First, we express the price (p) in terms of x:
\( 2p = 106 - x \)
\( p = \frac{106 - x}{2} \)
\( p = 53 - \frac{x}{2} \)
Next, we find the total revenue (R), which is given as \( R = px \):
\( R = (53 - \frac{x}{2})x \)
\( R = 53x - \frac{x^{2}}{2} \)
The average cost (AC) curve is given as \( AC = 5 + \frac{x}{50} \).
To find the total cost (C), we multiply AC by x:
\( C = AC \times x \)
\( C = (5 + \frac{x}{50})x \)
\( C = 5x + \frac{x^{2}}{50} \)
Now, we can find the profit function (P) by subtracting total cost from total revenue:
\( P = R - C \)
\( P = (53x - \frac{x^{2}}{2}) - (5x + \frac{x^{2}}{50}) \)
\( P = 53x - \frac{x^{2}}{2} - 5x - \frac{x^{2}}{50} \)
Combine like terms:
\( P = (53x - 5x) - (\frac{x^{2}}{2} + \frac{x^{2}}{50}) \)
\( P = 48x - (\frac{25x^{2}}{50} + \frac{x^{2}}{50}) \)
\( P = 48x - \frac{26x^{2}}{50} \)
\( P = 48x - \frac{13x^{2}}{25} \)
To find the most profitable output, we take the first derivative of the profit function and set it to zero:
\( \frac{d P}{d x} = \frac{d}{dx}(48x - \frac{13x^{2}}{25}) \)
\( \frac{d P}{d x} = 48 - \frac{13 \times 2x}{25} \)
\( \frac{d P}{d x} = 48 - \frac{26x}{25} \)
Set \( \frac{d P}{d x} = 0 \):
\( 48 - \frac{26x}{25} = 0 \)
\( 48 = \frac{26x}{25} \)
\( x = \frac{48 \times 25}{26} \)
\( x = \frac{1200}{26} \)
\( x = \frac{600}{13} \approx 46.1538 \)
Since the number of units must be a whole number, we round to \( x = 46 \) units.
To confirm this output yields maximum profit, we calculate the second derivative of the profit function:
\( \frac{d^{2} P}{d x^{2}} = \frac{d}{dx}(48 - \frac{26x}{25}) \)
\( \frac{d^{2} P}{d x^{2}} = - \frac{26}{25} \)
Since \( \frac{d^{2} P}{d x^{2}} \) is negative (less than 0), the profit is indeed maximized at \( x = 46 \) units. This derivative being negative confirms that the function is concave down at this point, indicating a maximum.
Now, we calculate the maximum profit by substituting \( x = 46 \) into the profit function:
\( P = 48(46) - \frac{13(46)^{2}}{25} \)
\( P = 2208 - \frac{13 \times 2116}{25} \)
\( P = 2208 - \frac{27508}{25} \)
\( P = 2208 - 1100.32 \)
\( P = \text{Rs. } 1107.68 \)
So, the most profitable output is 46 units, and the maximum profit is Rs. 1107.68.
In simple words: First, we find out how much money we make from selling items and how much it costs to make them. We put these together to get a profit formula. Then, we use special math (derivatives) to find the exact number of items to sell for the most profit. We found that selling 46 items gives the highest profit, which is Rs. 1107.68.

🎯 Exam Tip: Remember to express both revenue and cost in terms of a single variable (usually x) before forming the profit function, then use the first and second derivative tests to find and confirm maximum profit.

Question 4. A tour operator charges Rs. 136 per passenger with a discount of 40 paise for each passenger in excess of 100. The operator requires at least 100 passengers to operate the tour. Determine the number of passengers that will maximize the amount of money the tour operator receives.
Answer: Let x be the number of passengers. We are given that \( x \geq 100 \).
The base charge per passenger is Rs. 136.
The discount is 40 paise per passenger for each passenger in excess of 100. This means for every passenger over 100, the total price per passenger is reduced by 40 paise.
40 paise is equal to Rs. \( \frac{40}{100} = \frac{2}{5} \).
The number of passengers in excess of 100 is \( (x - 100) \).
The total discount applied to the price per passenger is \( \frac{2}{5}(x - 100) \).
So, the charge per passenger will be:
\( \text{Charge per passenger} = 136 - \frac{2}{5}(x - 100) \)
\( \text{Charge per passenger} = 136 - \frac{2x}{5} + \frac{2}{5} \times 100 \)
\( \text{Charge per passenger} = 136 - \frac{2x}{5} + 40 \)
\( \text{Charge per passenger} = 176 - \frac{2x}{5} \)
The total amount of money (A) the tour operator receives is the number of passengers multiplied by the charge per passenger:
\( A = x \times (\text{Charge per passenger}) \)
\( A = x(176 - \frac{2x}{5}) \)
\( A = 176x - \frac{2x^{2}}{5} \)
To maximize this amount, we find the first derivative of A with respect to x and set it to zero:
\( \frac{dA}{dx} = \frac{d}{dx}(176x - \frac{2x^{2}}{5}) \)
\( \frac{dA}{dx} = 176 - \frac{2 \times 2x}{5} \)
\( \frac{dA}{dx} = 176 - \frac{4x}{5} \)
Set \( \frac{dA}{dx} = 0 \):
\( 176 - \frac{4x}{5} = 0 \)
\( 176 = \frac{4x}{5} \)
\( 4x = 176 \times 5 \)
\( x = \frac{176 \times 5}{4} \)
\( x = 44 \times 5 \)
\( x = 220 \)
To confirm this is a maximum, we find the second derivative of A:
\( \frac{d^{2}A}{dx^{2}} = \frac{d}{dx}(176 - \frac{4x}{5}) \)
\( \frac{d^{2}A}{dx^{2}} = - \frac{4}{5} \)
Since \( \frac{d^{2}A}{dx^{2}} \) is negative (less than 0), the amount of money is maximized when the number of passengers is 220. This indicates the function is concave down at this point, confirming a peak.
In simple words: We figure out how the total money the operator gets changes with the number of passengers, considering the discount. We create a formula for this total money. Then, using a math tool called derivatives, we find the best number of passengers that will bring in the most money. The calculation shows that 220 passengers will give the operator the highest earnings.

🎯 Exam Tip: Be careful when formulating the revenue or charge function with discounts; ensure the discount is applied correctly, whether per passenger or to the total fare.

Question 5. Find the local minimum and local maximum of \( y = 2x^{3} - 3x^{2} - 36x + 10 \).
Answer: We are given the function \( y = 2x^{3} - 3x^{2} - 36x + 10 \).
First, we find the first derivative of y with respect to x to locate critical points:
\( \frac{dy}{dx} = \frac{d}{dx}(2x^{3} - 3x^{2} - 36x + 10) \)
\( \frac{dy}{dx} = 6x^{2} - 6x - 36 \)
Factor out 6:
\( \frac{dy}{dx} = 6(x^{2} - x - 6) \)
To find local minimum or maximum points, we set the first derivative to zero:
\( 6(x^{2} - x - 6) = 0 \)
\( x^{2} - x - 6 = 0 \)
Factor the quadratic equation:
\( (x - 3)(x + 2) = 0 \)
This gives us two critical points: \( x = 3 \) and \( x = -2 \).
Next, we find the second derivative of y to determine if these points are local minimums or maximums:
\( \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(6x^{2} - 6x - 36) \)
\( \frac{d^{2}y}{dx^{2}} = 12x - 6 \)
Now, we test each critical point:
**Case (i): When \( x = 3 \)**
Substitute \( x = 3 \) into the second derivative:
\( \left(\frac{d^{2}y}{dx^{2}}\right)_{x=3} = 12(3) - 6 \)
\( = 36 - 6 \)
\( = 30 \)
Since \( 30 \) is positive ( \( > 0 \) ), the function has a local minimum at \( x = 3 \).
To find the local minimum value, substitute \( x = 3 \) back into the original function:
\( y = 2(3)^{3} - 3(3)^{2} - 36(3) + 10 \)
\( y = 2(27) - 3(9) - 108 + 10 \)
\( y = 54 - 27 - 108 + 10 \)
\( y = 27 - 108 + 10 \)
\( y = -81 + 10 \)
\( y = -71 \)
So, the local minimum value is -71 at \( x = 3 \).
**Case (ii): When \( x = -2 \)**
Substitute \( x = -2 \) into the second derivative:
\( \left(\frac{d^{2}y}{dx^{2}}\right)_{x=-2} = 12(-2) - 6 \)
\( = -24 - 6 \)
\( = -30 \)
Since \( -30 \) is negative ( \( < 0 \) ), the function has a local maximum at \( x = -2 \).
To find the local maximum value, substitute \( x = -2 \) back into the original function:
\( y = 2(-2)^{3} - 3(-2)^{2} - 36(-2) + 10 \)
\( y = 2(-8) - 3(4) + 72 + 10 \)
\( y = -16 - 12 + 72 + 10 \)
\( y = -28 + 72 + 10 \)
\( y = 44 + 10 \)
\( y = 54 \)
So, the local maximum value is 54 at \( x = -2 \). Understanding these points helps visualize the shape of the graph, showing where it peaks and dips.
In simple words: We used math steps to find the highest and lowest points on the graph of the given equation. We found that the graph has a lowest point (minimum) of -71 when x is 3, and a highest point (maximum) of 54 when x is -2.

🎯 Exam Tip: Always remember to substitute the critical points back into the *original function* to find the local minimum or maximum values, not into the derivatives.

Question 6. The revenue function for a commodity is \( R = 15x + \frac{x^{2}}{3} - \frac{1}{36} x^{4} \). Show that at the highest point average revenue is equal to the marginal revenue.
Answer: We are given the total revenue function:
\( R = 15x + \frac{x^{2}}{3} - \frac{1}{36} x^{4} \)
First, we calculate the Average Revenue (AR) by dividing the total revenue by x:
\( AR = \frac{R}{x} = \frac{15x + \frac{x^{2}}{3} - \frac{1}{36} x^{4}}{x} \)
\( AR = 15 + \frac{x}{3} - \frac{1}{36} x^{3} \)
To find the highest point (maximum) of average revenue, we find the derivative of AR with respect to x and set it to zero:
\( \frac{d(AR)}{dx} = \frac{d}{dx}(15 + \frac{x}{3} - \frac{1}{36} x^{3}) \)
\( \frac{d(AR)}{dx} = 0 + \frac{1}{3} - \frac{3x^{2}}{36} \)
\( \frac{d(AR)}{dx} = \frac{1}{3} - \frac{x^{2}}{12} \)
Set \( \frac{d(AR)}{dx} = 0 \):
\( \frac{1}{3} - \frac{x^{2}}{12} = 0 \)
\( \frac{1}{3} = \frac{x^{2}}{12} \)
\( x^{2} = \frac{12}{3} \)
\( x^{2} = 4 \)
\( x = 2 \) (Since output x must be positive).
To confirm this x-value gives a maximum AR, we find the second derivative of AR:
\( \frac{d^{2}(AR)}{dx^{2}} = \frac{d}{dx}(\frac{1}{3} - \frac{x^{2}}{12}) \)
\( \frac{d^{2}(AR)}{dx^{2}} = 0 - \frac{2x}{12} \)
\( \frac{d^{2}(AR)}{dx^{2}} = - \frac{x}{6} \)
At \( x = 2 \):
\( \left(\frac{d^{2}(AR)}{dx^{2}}\right)_{x=2} = - \frac{2}{6} = - \frac{1}{3} \)
Since the second derivative is negative, AR is indeed maximum at \( x = 2 \).
Now, we calculate the Average Revenue at \( x = 2 \):
\( AR = 15 + \frac{2}{3} - \frac{1}{36} (2)^{3} \)
\( AR = 15 + \frac{2}{3} - \frac{8}{36} \)
\( AR = 15 + \frac{2}{3} - \frac{2}{9} \)
To combine these, find a common denominator (9):
\( AR = 15 + \frac{2 \times 3}{3 \times 3} - \frac{2}{9} \)
\( AR = 15 + \frac{6}{9} - \frac{2}{9} \)
\( AR = 15 + \frac{4}{9} \) ......... (1)
Next, we calculate the Marginal Revenue (MR) by finding the derivative of the total revenue function R with respect to x:
\( MR = \frac{dR}{dx} = \frac{d}{dx}(15x + \frac{x^{2}}{3} - \frac{1}{36} x^{4}) \)
\( MR = 15 + \frac{2x}{3} - \frac{4x^{3}}{36} \)
\( MR = 15 + \frac{2x}{3} - \frac{x^{3}}{9} \)
Now, we calculate the Marginal Revenue at \( x = 2 \):
\( MR = 15 + \frac{2(2)}{3} - \frac{(2)^{3}}{9} \)
\( MR = 15 + \frac{4}{3} - \frac{8}{9} \)
To combine these, find a common denominator (9):
\( MR = 15 + \frac{4 \times 3}{3 \times 3} - \frac{8}{9} \)
\( MR = 15 + \frac{12}{9} - \frac{8}{9} \)
\( MR = 15 + \frac{4}{9} \) ......... (2)
From equations (1) and (2), we can see that at \( x = 2 \) (the output where average revenue is highest), the Average Revenue (AR) is equal to the Marginal Revenue (MR). This illustrates a fundamental principle in economics where the marginal curve intersects the average curve at its maximum point.
In simple words: We found a formula for the average money earned per item and another for the extra money earned from one more item (marginal revenue). We then found the point where the average money per item was highest. At that exact point, we proved that the average money per item was the same as the extra money from one more item.

🎯 Exam Tip: Remember the economic principle that the marginal revenue (MR) curve intersects the average revenue (AR) curve at the point where AR is at its maximum. This means at \( AR_{max} \), \( MR = AR \).