Samacheer Kalvi Class 11 Business Maths Solutions Chapter 6 Applications of Differentiation Exercise 6.5

Question 1. Find the marginal productivities of capital (K) and labour (L) if \( P = 8L - 2K + 3K^2 - 2L^2 + 7KL \) when \( K = 3 \) and \( L = 1 \).
Answer: The given production function is \( P = 8L - 2K + 3K^2 - 2L^2 + 7KL \).
First, we find the marginal productivity of labour, which is the partial derivative of P with respect to L, treating K as a constant.
\( \frac{\partial P}{\partial L} = \frac{\partial}{\partial L} (8L - 2K + 3K^2 - 2L^2 + 7KL) \)
\( = 8 - 0 + 0 - 2(2L) + 7K(1) \)
\( = 8 - 4L + 7K \)
Now, we substitute the values \( K = 3 \) and \( L = 1 \) into this expression:
\( \left(\frac{\partial P}{\partial L}\right)_{(3,1)} = 8 - 4(1) + 7(3) \)
\( = 8 - 4 + 21 \)
\( = 4 + 21 \)
\( = 25 \)
Next, we find the marginal productivity of capital, which is the partial derivative of P with respect to K, treating L as a constant.
\( \frac{\partial P}{\partial K} = \frac{\partial}{\partial K} (8L - 2K + 3K^2 - 2L^2 + 7KL) \)
\( = 0 - 2(1) + 3(2K) - 0 + 7L(1) \)
\( = -2 + 6K + 7L \)
Finally, we substitute the values \( K = 3 \) and \( L = 1 \) into this expression:
\( \left(\frac{\partial P}{\partial K}\right)_{(3,1)} = -2 + 6(3) + 7(1) \)
\( = -2 + 18 + 7 \)
\( = 16 + 7 \)
\( = 23 \)
Marginal productivity helps businesses understand how adding one more unit of labor or capital affects their total output.
In simple words: We find how much the production changes if we add a little more labour, keeping capital the same. Then, we find how much it changes if we add a little more capital, keeping labour the same. We do this by using special math called partial derivatives and then put in the given numbers for capital and labour.

🎯 Exam Tip: Remember to treat the other variable as a constant when performing partial differentiation. Clearly show each step of substitution for full marks.

Question 2. If the production of a firm is given by \( P = 4LK - L^2 + K^2 \), \( L > 0 \), \( K > 0 \), Prove that \( L \frac{\partial P}{\partial L} + K \frac{\partial P}{\partial K} = 2P \).
Answer: The production function is given by \( P = 4LK - L^2 + K^2 \).
First, we find the partial derivative of P with respect to L:
\( \frac{\partial P}{\partial L} = \frac{\partial}{\partial L} (4LK - L^2 + K^2) \)
\( = 4K - 2L \)
Next, we find the partial derivative of P with respect to K:
\( \frac{\partial P}{\partial K} = \frac{\partial}{\partial K} (4LK - L^2 + K^2) \)
\( = 4L + 2K \)
Now, we substitute these into the expression \( L \frac{\partial P}{\partial L} + K \frac{\partial P}{\partial K} \):
\( L \frac{\partial P}{\partial L} + K \frac{\partial P}{\partial K} = L(4K - 2L) + K(4L + 2K) \)
\( = 4LK - 2L^2 + 4LK + 2K^2 \)
\( = 8LK - 2L^2 + 2K^2 \)
We also know that \( 2P = 2(4LK - L^2 + K^2) \)
\( = 8LK - 2L^2 + 2K^2 \)
Since \( L \frac{\partial P}{\partial L} + K \frac{\partial P}{\partial K} = 8LK - 2L^2 + 2K^2 \) and \( 2P = 8LK - 2L^2 + 2K^2 \), we have proven that:
\( L \frac{\partial P}{\partial L} + K \frac{\partial P}{\partial K} = 2P \)
This relationship is an example of Euler's Theorem for homogeneous functions, where P is a homogeneous function of degree 2.
In simple words: We calculated how much production changes when we change labour, and how much it changes when we change capital. Then we multiplied these changes by L and K, and added them up. We found that this sum was exactly two times the original production.

🎯 Exam Tip: For Euler's theorem proofs, clearly show each partial derivative and then substitute them into the theorem's formula. Ensure all terms combine correctly to match 2P.

Question 3. If the production function is \( z = 3x^2 - 4xy + 3y^2 \) where x is the labour and y is the capital, find the marginal productivities of x and y when \( x = 1, y = 2 \).
Answer: The given production function is \( z = 3x^2 - 4xy + 3y^2 \).
First, we find the marginal productivity of labour (x), which is the partial derivative of z with respect to x:
\( \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (3x^2 - 4xy + 3y^2) \)
\( = 6x - 4y + 0 \)
\( = 6x - 4y \)
Now, we substitute the values \( x = 1 \) and \( y = 2 \):
\( \left(\frac{\partial z}{\partial x}\right)_{(1,2)} = 6(1) - 4(2) \)
\( = 6 - 8 \)
\( = -2 \)
Next, we find the marginal productivity of capital (y), which is the partial derivative of z with respect to y:
\( \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (3x^2 - 4xy + 3y^2) \)
\( = 0 - 4x(1) + 3(2y) \)
\( = -4x + 6y \)
Finally, we substitute the values \( x = 1 \) and \( y = 2 \):
\( \left(\frac{\partial z}{\partial y}\right)_{(1,2)} = -4(1) + 6(2) \)
\( = -4 + 12 \)
\( = 8 \)
A negative marginal productivity of labour here suggests that adding more labour beyond this point might decrease overall output, which is an important insight for resource allocation.
In simple words: We calculate how much the total production (z) changes for a small change in labour (x) and then for a small change in capital (y). We use the given numbers for x and y to find these change rates.

🎯 Exam Tip: Pay close attention to the signs when performing partial differentiation, especially with terms like \( -4xy \). Ensure correct substitution of x and y values for both derivatives.

Question 4. For the production function \( P = 3(L)^{0.4} (K)^{0.6} \), find the marginal productivities of labour (L) and capital (K) when \( L = 10 \) and \( K = 6 \). [use: \( (0.6)^{0.6} = 0.736 \), \( (1.67)^{0.4} = 1.2267 \)]
Answer: The given production function is \( P = 3L^{0.4} K^{0.6} \).
First, we find the marginal productivity of labour, \( \frac{\partial P}{\partial L} \):
\( \frac{\partial P}{\partial L} = \frac{\partial}{\partial L} (3L^{0.4} K^{0.6}) \)
\( = 3K^{0.6} \frac{\partial}{\partial L} (L^{0.4}) \)
\( = 3K^{0.6} (0.4) L^{0.4-1} \)
\( = 1.2 K^{0.6} L^{-0.6} \)
\( = 1.2 \frac{K^{0.6}}{L^{0.6}} \)
\( = 1.2 \left(\frac{K}{L}\right)^{0.6} \)
Now, we substitute \( L = 10 \) and \( K = 6 \):
\( \frac{\partial P}{\partial L} = 1.2 \left(\frac{6}{10}\right)^{0.6} \)
\( = 1.2 (0.6)^{0.6} \)
Using the given value \( (0.6)^{0.6} = 0.736 \):
\( = 1.2 \times 0.736 \)
\( = 0.8832 \)
Next, we find the marginal productivity of capital, \( \frac{\partial P}{\partial K} \):
\( \frac{\partial P}{\partial K} = \frac{\partial}{\partial K} (3L^{0.4} K^{0.6}) \)
\( = 3L^{0.4} \frac{\partial}{\partial K} (K^{0.6}) \)
\( = 3L^{0.4} (0.6) K^{0.6-1} \)
\( = 1.8 L^{0.4} K^{-0.4} \)
\( = 1.8 \frac{L^{0.4}}{K^{0.4}} \)
\( = 1.8 \left(\frac{L}{K}\right)^{0.4} \)
Now, we substitute \( L = 10 \) and \( K = 6 \):
\( \frac{\partial P}{\partial K} = 1.8 \left(\frac{10}{6}\right)^{0.4} \)
\( = 1.8 (1.66666)^{0.4} \)
Using the given value \( (1.67)^{0.4} = 1.2267 \) (approximating 1.66666 to 1.67):
\( = 1.8 \times 1.2267 \)
\( = 2.20806 \)
\( \approx 2.2081 \)
This type of production function, known as a Cobb-Douglas function, is widely used in economics because it shows diminishing marginal returns to scale.
In simple words: We calculate how much extra production we get by adding one more unit of labour, and then one more unit of capital. We use special math rules for powers and then plug in the given numbers for labour and capital to find the exact values.

🎯 Exam Tip: When dealing with fractional exponents, remember the power rule \( \frac{d}{dx} (x^n) = nx^{n-1} \). Be careful with negative exponents when simplifying and ensure you use the provided values for calculations accurately.

Question 5. The quantity A is \( q = 13 - 2p_1 - 3p_2^2 \). Find the partial elasticities \( \frac{E_q}{E_{p_1}} \) and \( \frac{E_q}{E_{p_2}} \) when \( p_1 = 2 \) and \( p_2 = 2 \).
Answer: The quantity function is given by \( q = 13 - 2p_1 - 3p_2^2 \).
First, we find the partial derivative of q with respect to \( p_1 \):
\( \frac{\partial q}{\partial p_1} = \frac{\partial}{\partial p_1} (13 - 2p_1 - 3p_2^2) \)
\( = 0 - 2(1) - 0 \)
\( = -2 \)
Now, we calculate the partial elasticity of q with respect to \( p_1 \), denoted as \( E_{q,p_1} \):
\( E_{q,p_1} = \frac{p_1}{q} \frac{\partial q}{\partial p_1} \)
We need to find q when \( p_1 = 2 \) and \( p_2 = 2 \):
\( q = 13 - 2(2) - 3(2)^2 \)
\( = 13 - 4 - 3(4) \)
\( = 13 - 4 - 12 \)
\( = 9 - 12 \)
\( = -3 \)
Substitute the values into the elasticity formula:
\( E_{q,p_1} = \frac{2}{-3} (-2) \)
\( = \frac{4}{3} \)
Next, we find the partial derivative of q with respect to \( p_2 \):
\( \frac{\partial q}{\partial p_2} = \frac{\partial}{\partial p_2} (13 - 2p_1 - 3p_2^2) \)
\( = 0 - 0 - 3(2p_2) \)
\( = -6p_2 \)
Now, we calculate the partial elasticity of q with respect to \( p_2 \), denoted as \( E_{q,p_2} \):
\( E_{q,p_2} = \frac{p_2}{q} \frac{\partial q}{\partial p_2} \)
We already found \( q = -3 \) when \( p_1 = 2 \) and \( p_2 = 2 \).
Substitute the values into the elasticity formula:
\( E_{q,p_2} = \frac{2}{-3} (-6 \times 2) \)
\( = \frac{2}{-3} (-12) \)
\( = \frac{-24}{-3} \)
\( = 8 \)
Elasticity helps us understand how sensitive the quantity demanded is to changes in price, which is crucial for pricing strategies.
In simple words: We figure out how much the quantity changes if the price of item 1 goes up a little, and then how much it changes if the price of item 2 goes up a little. We use the current prices to calculate these "sensitivities."

🎯 Exam Tip: Always calculate the value of q first for the given prices, as it is needed in both elasticity formulas. Remember that partial elasticities measure sensitivity to one price while holding other prices constant.

Question 6. The demand for a commodity A is \( q = 80 - p_1^2 + 5p_2 - p_1p_2 \). Find the partial elasticities \( \frac{E_q}{E_{p_1}} \) and \( \frac{E_q}{E_{p_2}} \) when \( p_1 = 2, p_2 = 1 \).
Answer: The demand function is given by \( q = 80 - p_1^2 + 5p_2 - p_1p_2 \).
First, we find the partial derivative of q with respect to \( p_1 \):
\( \frac{\partial q}{\partial p_1} = \frac{\partial}{\partial p_1} (80 - p_1^2 + 5p_2 - p_1p_2) \)
\( = 0 - 2p_1 + 0 - (1)p_2 \)
\( = -2p_1 - p_2 \)
Now, we calculate the partial elasticity of q with respect to \( p_1 \), denoted as \( E_{q,p_1} \):
\( E_{q,p_1} = \frac{p_1}{q} \frac{\partial q}{\partial p_1} \)
We need to find q when \( p_1 = 2 \) and \( p_2 = 1 \):
\( q = 80 - (2)^2 + 5(1) - (2)(1) \)
\( = 80 - 4 + 5 - 2 \)
\( = 76 + 5 - 2 \)
\( = 81 - 2 \)
\( = 79 \)
Substitute the values into the elasticity formula:
\( E_{q,p_1} = \frac{2}{79} (-2(2) - 1) \)
\( = \frac{2}{79} (-4 - 1) \)
\( = \frac{2}{79} (-5) \)
\( = -\frac{10}{79} \)
Next, we find the partial derivative of q with respect to \( p_2 \):
\( \frac{\partial q}{\partial p_2} = \frac{\partial}{\partial p_2} (80 - p_1^2 + 5p_2 - p_1p_2) \)
\( = 0 - 0 + 5(1) - p_1(1) \)
\( = 5 - p_1 \)
Now, we calculate the partial elasticity of q with respect to \( p_2 \), denoted as \( E_{q,p_2} \):
\( E_{q,p_2} = \frac{p_2}{q} \frac{\partial q}{\partial p_2} \)
We already found \( q = 79 \) when \( p_1 = 2 \) and \( p_2 = 1 \).
Substitute the values into the elasticity formula:
\( E_{q,p_2} = \frac{1}{79} (5 - 2) \)
\( = \frac{1}{79} (3) \)
\( = \frac{3}{79} \)
These elasticity values tell us how much the demand for commodity A changes when the price of \( p_1 \) or \( p_2 \) changes, which is vital for businesses to set prices and understand market reactions.
In simple words: We calculate how sensitive the demand for an item is to changes in its own price and also to changes in the price of another related item. We use the current demand and prices to find these 'sensitivity numbers'.

🎯 Exam Tip: Remember to calculate q using the given \( p_1 \) and \( p_2 \) values before finding the elasticities. Pay careful attention to the signs and algebraic manipulation when computing the partial derivatives.