Q2 Answer The partial derivatives of the function \(f(x_1,x_2 )=ax_1+bx_2\) are $${{∂f(x_1,x_2 )}\over {∂x_1}}=f_1=a$$ $${{∂f(x_1,x_2 )}\over {∂x_2}}=f_2=b$$

Note:
When we compute \({{∂f(x_1,x_2 )}\over {∂x_1}}\), the second term \(bx_2\) in the function appears as an additive constant, and therefore, after the partial differentiation, the second term disappears.

Similarly, when we compute \({{∂f(x_1,x_2 )}\over {∂x_2}}\), the first term \(ax_1\) in the function appears as the additive constant, and differentiation of the first term is, therefore, zero. Show Question

Q3 Answer The partial derivatives of the revenue function \(f(x_1,x_2 )= p_1 x_1+p_2 x_2\) are $${{∂f(x_1,x_2 )}\over {∂x_1}}=f_1=p_1$$ $${{∂f(x_1,x_2 )}\over {∂x_2}}=f_2=p_2$$

Partial differentiation of the revenue function \(f(x_1,x_2,…,x_n )\) with respect to one good \(x_i\) is called the marginal revenue from good \(i\). This is the amount the firm earns by selling an extra unit of good i holding the amount of other goods fixed. In this example the marginal revenue of good \(i\) is \(p_i\), the price of the good.

This is an example from a competitive market where the marginal revenue of a good is the market price of the good. Firms operate in non-competitive markets generally face downward-sloping marginal revenue curves. Show Question

Q4 Answer The partial derivatives of the function \(f(x_1,x_2 )=x_1 x_2\) are $${{∂f(x_1,x_2 )}\over {∂x_1}}=f_1=x_2$$ $${{∂f(x_1,x_2 )}\over {∂x_2}}=f_2=x_1$$

Note:
The variable \(x_j\) is held fixed when computing \({{∂f}\over {∂x_i}}\). However, the variable \(x_j\) is a multiplicative constant in this function, and therefore, reappears after differentiation. Show Question

Q5 Answer The partial derivatives of the function \(f(x_1,x_2 )=x_1^{0.5} x_2^{0.5}\) are $${{∂f(x_1,x_2 )}\over {∂x_1}}=f_1=0.5 x_1^{-0.5} x_2^{0.5}$$ $${{∂f(x_1,x_2 )}\over {∂x_2}}=f_2=0.5 x_1^{0.5} x_2^{-0.5}$$

Note:
Apply the power rule of differentiation of one variable function when computing \({{∂f}\over{∂x_i}}\) by holding \(x_j\) as a multiplicative constant.

The function \(f(x_1,x_2 )=x_1^{0.5} x_2^{0.5}\) is an example of one of the most frequently used Cobb-Douglas functions in economics. To read more about the Cobb-Douglas functions, use recommended books. Show Question

Q6 Answer The partial derivatives of the function \(f(x_1,x_2 )=Ax_1^α x_2^β\) are $${{∂f(x_1,x_2 )}\over {∂x_1}}=f_1=Aαx_1^{α-1} x_2^β$$ $${{∂f(x_1,x_2 )}\over {∂x_2}}=f_2=Aβx_1^α x_2^{β-1}$$

Note:
Apply the power rule of differentiation of one variable function when computing \({{∂f}\over {∂x_i}}\) by holding \(x_j\) fixed. See that A is also a multiplicative constant in the function, hence, does not disappear after the partial differentiation. Show Question

Q7 Answer The partial derivatives of the production function \(f(x_1,x_2 )=12x_1^{0.5} x_2^{0.5}\) are $${{∂f(x_1,x_2 )}\over {∂x_1}}=f_1=6 x_1^{-0.5} x_2^{0.5}$$ $${{∂f(x_1,x_2 )}\over {∂x_2}}=f_2=6 x_1^{0.5} x_2^{-0.5}$$

Here, \(f_1\) and \(f_2\) represent the marginal product of input one (suppose, labor) and input two (suppose, capital), respectively.

Note (Optional):
In this example we see that, holding the amount of second input \((x_2)\) fixed, if the firm increases the amount of first input \((x_1)\), marginal productivity of input one decreases. On the other hand, if the amount of second input \((x_2)\) increases, marginal productivity of input one also increases.
Consider three pairs of inputs \((4, 4), (9, 4)\) and \((9, 9)\).
At \((4,4)\), the marginal product of input \(1\) is: \(f_1 (4,4)=6 x_1^{-0.5}x_2^{0.5}={{6x_2^{0.5}}\over x_1^{0.5}}=6\)
At \((9,4)\), the marginal product of input \(1\) is: \(f_1 (9,4)=6 x_1^{-0.5} x_2^{0.5}={{6x_2^{0.5}}\over x_1^{0.5}}=4\)
At \((9,9)\), the marginal product of input \(1\) is: \(f_1 (9.9)=6 x_1^{-0.5} x_2^{0.5}={{6x_2^{0.5}}\over x_1^{0.5}}=6\)
Similar findings are also observed for the productivity of second input in this example.

In general, for any Cobb-Douglas production function, an increase in input \(j\) increases the productivity of input \(i\), where \(i≠j\). Holding input \(j\) fixed, any increase in input \(i\) decreases the productivity of input \(i\). Show Question

Q8 Answer The partial derivatives of the function \(f(x_1,x_2 )=10x_1+6x_2\) are $$f_1=10$$ $$f_2=6$$ Along any level curve \(df(x_1,x_2 )=0\) and the slope of the level curve is $$ {{dx_2}\over {dx_1}}|_{d f(x_1,x_2 )=0}=-{{f_1}\over {f_2}} $$ The slope of the level curves for the function \(f(x_1,x_2 )=10x_1+6x_2\) is $$-{{f_1}\over {f_2}} = -{{10}\over {6}}=-{{5}\over 3}$$

Note: Consider a function \(f(x_1,x_2 )\). The set of \((x_1,x_2)\) pairs that generate a specific level for the function, for example, \(f(x_1,x_2 )=k\)   where   \(k\)   is a constant, is called the level set corresponds to the level \(k\). In economics, indifference curves and isoquants are examples of frequently used level sets.

If the given function is a function of two variables, and we can solve the corresponding equation for the level curve explicitly for \(x_2\) in terms of \(x_1\), then we have the equation for a level curve in \((x_1,x_2)\) Cartesian plane. When we can derive this equation explicitly for \(x_2\), the derivative \({{dx_2}\over dx_1}\) presents the slope of the level curve.

However, it may not always be easy to express an equation corresponds to a level curve explicitly. In such cases, the approach of total differential may be helpful to find the slope of the level curves. Note again that, within the level curve \(d f(x_1,x_2 )=0\) as the function is held constant at \(f(x_1,x_2 )=k\). Using the total differential approach when \(d f(x_1,x_2 )=0\), we can derive the slope of the level curves without solving the equation explicitly: $${{dx_2}\over {dx_1}}=-{{f_1}\over f_2}$$ Here, \(f_1\) and \(f_2\) are the partial derivatives of the function \(f(x_1,x_2)\).
The total differential approach is not discussed here but you can study the approach from any recommended book. Show Question

Q9 Answer The partial derivatives of the function \(f(x_1,x_2)=x_1^{0.5} x_2^{0.5}\) are $$f_1=0.5x_1^{-0.5} x_2^{0.5}$$ $$f_2=0.5x_1^{0.5} x_2^{-0.5}$$ The slope of the level curve is $$ {{dx_2}\over {dx_1}}|_{d f(x_1,x_2 )=0}=-{{f_1}\over f_2} = -{{0.5x_1^{-0.5} x_2^{0.5}}\over {0.5x_1^{0.5} x_2^{-0.5}}}=-{{x_2}\over x_1} $$ Or, simply $${{dx_2}\over {dx_1}}=-{{f_1}\over f_2} =-{{x_2}\over {x_1}}$$

Q10 Answer The partial derivatives of the function \(f(x_1,x_2 )=x_1^{0.6} x_2^{0.4}\) are $$f_1=0.6x_1^{-0.4} x_2^{0.4}$$ $$f_2=0.4x_1^{0.6} x_2^{-0.6}$$ The slope of the level curve is $$ {{dx_2}\over {dx_1}}=-{{f_1}\over f_2} = -{{0.6x_1^{-0.4} x_2^{0.4}}\over {0.4x_1^{0.6} x_2^{-0.6}}}=-{{3x_2}\over 2x_1} $$ Note:
The exponents of the variables in the function play an important role in determining the slope of the level curves. Show Question

Q11 Answer The partial derivatives of the function \(f(x_1,x_2 )=10x_1^{0.6} x_2^{0.4}\) are $$f_1=6x_1^{-0.4} x_2^{0.4}$$ $$f_2=4x_1^{0.6} x_2^{-0.6}$$ The slope of the level curve is $$ {{dx_2}\over {dx_1}}=-{{f_1}\over f_2} = -{{6x_1^{-0.4} x_2^{0.4}}\over {4x_1^{0.6} x_2^{-0.6}}}=-{{3x_2}\over 2x_1} $$ Note:
The coefficient of the variables in the Cobb-Douglas function, in this example \(10\), does not influence the slope of the level curves. Show Question

Q12 Answer The marginal rate of substitution (MRS) is:

the absolute value of the slope of the level curves, given that the level curves are downward sloping.

The marginal rate of substitution (MRS) for the function \(f(x_1,x_2 )=10x_1^{0.5} x_2^{0.5}\) is:

Note:
Marginal rate of substitution (MRS) is the rate at which one variable can be substituted for another holding the level of the function constant.
Consider a level curve corresponds to the function \(f(x_1,x_2 )\) where each pair of \((x_1,x_2 )\) always generate a fixed level for the function $$f(x_1,x_2 )=k$$ The MRS is the rate at which we can substitute one variable for the other to keep the level of function fixed at \(f(x_1,x_2 )=k\).
To sum up, in a level curve corresponds to \(f(x_1,x_2 )=k\)  :
-   the change in the value of the function is always zero, that is \(df(x_1,x_2 )=0\)
-   the slope of the level curve is \({{dx_2}\over dx_1}=-{{f_1}\over f_2}\)
-   and the MRS is the absolute value of the slope of the level curves, that is,

Q13 Answer The marginal rate of substitution (MRS) between \(x_1\) and \(x_2\) for the function \(f(x_1,x_2 )=Ax_1^α x_2^β\) is:

Note:
See that, \(A\), which shows the technological parameter of the function, does not influence the MRS.
The ratio of the exponents of \(x_1\) and \(x_2\) in the function plays an important role to determine the MRS. If the exponents are the same, the MRS is simply the ratio of two variables at the corresponding point of any level curve. Show Question

Q14 Answer The marginal rate of substitution (MRS) between \(x_1\) and \(x_2\) for the utility function \(u=2x_1^{0.5} x_2^{0.5}\) is:

Interpretation:
A level curve corresponds to the utility function \(f(x_1,x_2 )\) is called the indifference curve. By connecting the pair of the goods \((x_1,x_2)\) that generate a fixed level of utility, we get the indifference curve. Along an indifference curve, the change in utility is always zero (\(df(x_1,x_2 )=0\)), and the marginal rate of substitution is MRS\(=-{{dx_2}\over {dx_1}}\). In this example MRS\(=-{{dx_2}\over {dx_1}}= {{x_2}\over {x_1}} \)

Consider, a pair of goods \((4, 16)\), This pair of goods lies on the indifference curve where the level of utility is:

And, at this pair of the point \((4, 16)\) of the indifference curve, MRS\(={{x_2}\over {x_1}} ={{16}\over 4}=4\). which shows the rate at which the consumer is willing to substitute \(x_2\) for \(x_1\) to keep the level of satisfaction same at \(16\).     Show Question

Q15 Answer The marginal rate of substitution (MRS) is:

The pair of points \((16, 4)\) and \((4,16)\) are on the same indifference curve as for both pairs \(u=x_1^{0.5} x_2^{0.5}=8\). However, the marginal rate of substitution at \((16, 4)\) is \(0.25\) while at \((4,16)\) is \(4\). For a Cobb-Douglas indifference curve, the slope and marginal rate of substitution are different at different points.

Note:
Let us explain why in a Cobb-Douglas function the MRS is different at different point using the given points (consumption baskets) in the question.

Consider the point \((16, 4)\) on the indifference curve where the consumer consumes \(16\) units of \(x_1\) and only \(4\) units of \(x_2\). At this point the consumption basket already has relatively more \(x_1\). Thus, the consumer is less willing to substitute \(x_2\) for \(x_1\). She is willing to substitute only \(0.25\) unit of \(x_2\) for one additional unit of \(x_1\) to remain equally happy.

Now consider the point \((4, 16)\) which is also on the same indifference curve. This consumption basket has only \(4\) units of \(x_1\) but \(16\) units of \(x_2\); the basket has relatively more \(x_2\) and it is worthwhile for the consumer to substitute more \(x_2\) for \(x_1\). That is exactly what is evident from the MRS at \((4, 16)\); the consumer is willing to sacrifice \(4\) units of \(x_2\) for only one additional unit of \(x_1\) to keep the level of satisfaction same. Show Question

Q16 Answer The partial derivatives of the function \(f(x_1,x_2 )=10x_1+5x_2\) are $$f_1=10$$ $$f_2=5$$ The slope of the level curves correspond to the function \(f(x_1,x_2 )=10x_1+5x_2\) is $${{dx_2}\over dx_1}=-{{f_1}\over f_2} =-{{10}\over 5}=-2$$ The marginal rate of substitution is

We see that from any initial consumption bundle, such as \((4, 4)\), if the consumer substitutes \(2\) units of \(x_2\) for one unit of \(x_1\), she remains on the same indifference curve with a level of utility \(f(x_1,x_2 )=10x_1+5x_2=60\). For example, at (5,2) the level of utility is \(f(x_1,x_2 )=10(5)+5(2)=60\). Since the marginal rate of substitution for these two goods is always \(2\), we say that these two commodities are perfect substitutes.

Note:
The utility function corresponds to the perfect substitutes is: $$f(x_1,x_2 )=αx_1+βx_2 $$ Where \(α\) and \(β\) are constants. Show Question

Q17 Answer The partial derivatives of the function \(f(x_1,x_2 )=Ax_1^α x_2^β\) are $$f_1=Aαx_1^{α-1} x_2^β$$ $$f_2=Aβx_1^α x_2^{β-1}$$ The slope of the level curve is $${{dx_2}\over dx_1}=-{{f_1}\over f_2} = -{{Aαx_1^{α-1} x_2^β}\over {Aβx_1^α x_2^{β-1}}}=-{{αx_2}\over βx_1}$$ The marginal rate of technical substitution is

Note:
The MRTS shows the amount of \(x_2\) released for one more unit of \(x_1\) is employed while keeping the output constant. Like the MRS, the technical efficiency parameter \(A\) does not influence the marginal rate of technical substitution if the production function is a Cobb-Douglas function. The exponents of the factors of production play an important role in determining the marginal rate of substitution between the factors of production. Show Question

Q18 Answer
This is a special form of production function where two factors of production are perfect complements, and the production process requires a fixed proportion of the factors of production. In this production function, no substitution is possible between the factors of production. Think about an example where \(1\) molecule of water requires \(2\) atoms of hydrogen and \(1\) atom of oxygen; no substitution is possible between hydrogen and oxygen.

The level curves correspond to the production function where the factors of production are perfect complements are L-shaped and are not differentiable at the points where the level curves are kinked. For such isoquants (indifference curves), we cannot compute the marginal rate of technical substitution (marginal rate of substitution) using the method of differentiation. If the level curve is L-shaped, the MRS is zero at the horizontal portion (excluding the kink point), infinity at the vertical portion (excluding the kink), and not defined at the kink point.

Consider an example where

In this example, to produce \(20\) units of output, \(10\) units of input one and \(20\) units of input two are always required and substitution is not possible at all at the efficient point of production. Show Question

Q19 Answer
Given the function to be optimized $$f(x_1,x_2 )=2x_1^2-x_1 x_2+3x_2^2-14x_1-8x_2$$ Take the first-order partial derivatives and set them equal to zero:

Using the method of elimination or substitution (or graphical approach, matrix inversion, whatever appropriate method you like), solve these two equations simultaneously to get the critical point at which the function may be optimized. Here, the critical point is \((4, 2)\).

Note:
1.   A necessary condition for the existence of a critical point in a multivariable function is that all first-order partial derivatives must be simultaneously \(0\).

2.   Without checking the second-order conditions, we cannot confirm that a function is optimized at a critical point. To read second-order conditions of optimization, please use recommended books Show Question

Q20 Answer
Here the problem is to maximize: \(f(x_1,x_2 )=x_1^{0.5} x_2^{0.5}\)
Subject to the constraint: \(g(x_1,x_2 )≡x_1+4x_2=120\)
The slope of the level curves correspond to the utility function is $${{dx_2}\over dx_1}=-{{f_1}\over f_2} =-{{x_2}\over x_1}$$ The slope of the constraint equation is $$-{{g_1}\over g_2} =-{{1}\over 4}$$ Finding the critical point using the geometric approach requires that the critical point is an interior point and at that point the slope of the level curve and the slope of the equality constraint are equal:

The necessary condition to be the critical point of the function in this example is that $$x_1=4x_2$$ Using this relation in the constraint function, we get

The critical point is \((60, 15)\) where the function may be optimized subject to the constraint. Show Question

Q21 Answer
The Lagrange function is $$L(x_1,x_2,μ)=x_1x_2-μ(2x_1+4x_2-64)$$ Set its partial derivatives equal to zero

From equations \((1)\) and \((2)\) we find: $$x_1=2x_2$$ Substitute \(x_1=2x_2\) in equation \((3)\)

Critical point is \((16, 8)\).

Note:
1.   Geometric approach and the Lagrange method to get optimal values work when there exists interior optimum points like the cases of typical utility maximization problems or cost minimization problems in economics where the objective function takes the form of Cobb-Douglas functions and the constraint function is a linear function or vice versa.

2.   In an interior optimum, at the optimal point the constraint function is tangent to the objective function. Show Question

Q22 Answer The Lagrange function is $$L(x_1,x_2,μ)=x_1^α x_2^{1-α}-μ(P_1 x_1+P_2 x_2-m)$$ Set its partial derivatives equal to zero

Rearrange and divide equation \((1)\) by \((2)\):

Use equation \((4)\) in equation \((3)\) to get:

Using \(x_2={{(1-α)m}\over P_2}\) in equation \((4)\) we get $$x_1={{αm}\over P_1}$$ We see that the demand for good \(i\) is inversely related with its own price and positively related with the income. Show Question

Q23 Answer
The optimization problem is to minimize: \(f(x_1,x_2 )=x_1+4x_2\)
Subject to the constraint: \(g(x_1,x_2 )≡8x_1^{0.5} x_2^{0.5}=64\)
The slope of the objective function $$-{{f_1}\over f_2} =-{{1}\over 4}$$ The slope of the constraint equation is $$-{{g_1}\over g_2} =-{{x_2}\over x_1}$$ Finding the critical point using the geometric approach requires that the critical point is an interior point which is equivalent to: $$-{{f_1}\over f_2} =-{{g_1}\over g_2}$$ $$-{{1}\over 4}=-{{x_2}\over x_1}$$ $$x_1=4x_2$$ Using this relation in the constraint function, we get

The objective function may be minimized at \((16,4)\) Show Question

Q24 Answer Here the problem is to minimize the cost:

Subject to the constraint $$g(x_1,x_2 )≡2x_1 x_2=256$$ As the constraint function is a Cobb-Douglas function while the objective function is a linear function, the critical point where the cost may be minimized is an interior point. Thus, at the optimal point where the cost is minimized $$-{{2}\over 4}=-{{x_2}\over x_1} $$ $$x_1=2x_2$$ Using \(x_1=2x_2\) in the constraint function we get $$2x_1 x_2=256$$ $$2(2x_2)x_2=256$$ $$x_2=8$$ Using \(𝑥_2=8\) we get that the firm will use \(𝑥_1=2𝑥_2=16\) units of labor to minimize the cost. Show Question

Q25 Answer Here the constrained optimization problem is to maximize utility: $$f(x_1,x_2)=2x_1^{0.6} x_2^{0.4}$$ Subject to the constraint $$g(x_1,x_2 )≡40x_1+20x_2=600$$ As the objective function is a Cobb-Douglas function while the constraint function is a linear function, the critical point where the utility may be maximized is an interior point. Thus, at the optimal point

Using \(x_1={{3}\over 4} x_2\) in the constraint function we get

Using \(𝑥_2=12\) we get \(𝑥_1={{3}\over 4}𝑥_2=9\) Show Question

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