1. Today’s Class… Derivatives of Multivariate Functions
Total Differentiation of Multivariate Functions
Unconstrained Optimization
Constrained Optimization using the Lagrangian
Constrained Optimization of Economic Functions

2. Derivatives of multivariate functions

3. Multivariate Functions: Definition
Our study of derivatives so far has been limited to functions of single independent variables such as y=f(x).
However, many economics activities involve functions with more than one independent variable
Z= f(x,y) is defined as a function of two independent variables if there exists one and only one value of z in the range of f for each ordered pair of real numbers (x,y) in the domain of f

4. PARTIAL DERIVATIVES
To measure the effect of a change in a single independent variable (x or y) on the dependent variable (z) in a multivariate function, the partial derivative is required.
The partial derivative of z with respect to x (y) measures the instantaneous rate of change of z with respect to x(y) while y(x) is held constant.
The partial derivative of z or f with respect to x (y) is denoted by the following notations:
𝜕𝑧 𝜕𝑥 , 𝜕𝑓 𝑑𝑥 , 𝑓 𝑥 (x,y), 𝑓 𝑥 or 𝑧 𝑥 and
𝜕𝑧 𝜕𝑦 , 𝜕𝑓 𝑑𝑦 , 𝑓 𝑦 (x,y), 𝑓 𝑦 or 𝑧 𝑦

5. Partial Derivatives: Mathematical Expression
The partial derivative of a multivariate function can be expressed mathematically as:
Rule for finding the partial differentiation of z = f(x,y):
To find 𝑓 𝑥 , regard y as a constant and differentiate f (x, y) with respect to x.
To find 𝑓 𝑦 , regard x as a constant and differentiate f (x, y) with respect to y.

6. Partial Derivatives: Examples
Find the partial derivative of the multivariate function: Z= 3 𝑥 2 𝑦 3
Solution:
Z= 3 𝑥 2 𝑦 3
When differentiating with respect to x, treat the y term as a constant by mentally bracketing it with the coefficient: Z= (3 𝑦 3 ). 𝑥 2
Then take the derivative of the x term holding the y term constant:
𝜕𝑧 𝜕𝑥 = 𝑧 𝑥 = (3 𝑦 3 ). 2x = 6x 𝑦 3
When differentiation with respect to y, treat the x term as constant by mentally bracketing it with the coefficient: Z= (3 𝑥 2 ).𝑦 3
Then take the derivative of the y term holding the x term constant:
𝜕𝑧 𝜕𝑦 = 𝑧 𝑥 = (3 𝑥 2 ). 3 𝑦 2 = 9 𝑥 2 𝑦 2

7. Trial Question
Take a few minutes to find the partial derivatives of the function
Z= 5 𝑥 𝑥 2 𝑦 2 +7 𝑦 5 , by following the steps in the immediate previous slide.

8. Partial Derivatives: Rules
Partial derivatives follow the same basic patterns as the rules of differentiation discussed in the earlier classes. The rules are as follows:
The Generalized Power Function Rule:
Given z= [ 𝑔(𝑥,𝑦)] 𝑛 : The partial derivatives with respect to x and y respectively are:
𝜕𝑧 𝜕𝑥 = n [ 𝑔(𝑥,𝑦)] 𝑛−1 . 𝜕𝑔 𝜕𝑥
Example: Differentiate z= ( 𝑥 3 +7 𝑦 2 ) 4 with respect to x and y.
𝜕𝑧 𝜕𝑥 = 4 ( 𝑥 3 +7 𝑦 2 ) 3 .(3 𝑥 2 )=12 𝑥 2 ( 𝑥 3 +7 𝑦 2 ) 3
𝜕𝑧 𝜕𝑦 = 4( 𝑥 3 +7 𝑦 2 ) 3 .(14y)=56y ( 𝑥 3 +7 𝑦 2 ) 3

9. Partial Derivatives: Rules
The Product Rule:
Given z= g(x,y) x h (x,y).The partial derivatives with respect to x and y are:
𝜕𝑧 𝜕𝑥 =g(x,y). 𝜕ℎ 𝜕𝑥 + h(x,y). 𝜕𝑔 𝜕𝑥
𝜕𝑧 𝜕𝑦 =g(x,y). 𝜕ℎ 𝜕𝑦 + h(x,y). 𝜕𝑔 𝜕𝑦
Example: Differentiate z= (3x+5)(2x+6y)with respect to x and y.
𝜕𝑧 𝜕𝑥 = (3x+5).(2)+(2x+6y)(3)=12x+10-18y
𝜕𝑧 𝜕𝑦 =(3x+5).(6)+(2x+6y).(0)= 18x+30

10. Partial Derivatives: Rules
The Quotient Rule:
Given z= g(x,y) / h (x,y).The partial derivatives with respect to x and y are:
𝜕𝑧 𝜕𝑥 = ℎ 𝑥,𝑦 . 𝜕𝑔 𝑑𝑥 − 𝑔(𝑥,𝑦) 𝜕ℎ 𝜕𝑥 [ ℎ(𝑥,𝑦)] 2
𝜕𝑧 𝜕𝑦 = ℎ 𝑥,𝑦 . 𝜕𝑔 𝑑𝑦 − 𝑔(𝑥,𝑦) 𝜕ℎ 𝜕𝑦 [ ℎ(𝑥,𝑦)] 2
Example: Differentiate z= (6x+7y)(5x+3y)with respect to x and y.
𝜕𝑧 𝜕𝑥 = 5𝑥+3𝑦 . 6 − 6𝑥+7𝑦 . (5) [ (5𝑥+3𝑦)] 2 = 30𝑥+18𝑦−30𝑥−35𝑦 [ 5𝑥+3𝑦] 2 = −17𝑦 (5𝑥+3𝑦) 2
𝜕𝑧 𝜕𝑦 = 5𝑥+3𝑦 . 7 − 6𝑥+7𝑦 . (3) [ (5𝑥+3𝑦)] 2 = 35𝑥+21𝑦−18𝑥−21𝑦 [ 5𝑥+3𝑦] 2 = 17𝑥 (5𝑥+3𝑦) 2

11. Second Order Partial Derivatives
Given a function z= f(x,y), the direct second order partial derivative signifies that the function has been differentiated with respect to one of the independent variables twice while the other independent variable has been held constant.
This is denoted by 𝑓 𝑥𝑥 = 𝜕 𝜕𝑥 𝜕𝑧 𝜕𝑥 = 𝜕 2 𝑧 𝜕 𝑥 and 𝑓 𝑦𝑦 = 𝜕 𝜕𝑦 𝜕𝑧 𝜕𝑦 = 𝜕 2 𝑧 𝜕 𝑦 2
𝑓 𝑥𝑥 measures the rate of change of the first-order partial derivative 𝑓 𝑥 with respect to x while y is held constant
𝑓 𝑦𝑦 measures the rate of change of the first –order partial derivative 𝑓 𝑦 with respect to y while x is held constant

12. Second Order Partial Derivatives
The cross or mixed partial derivatives 𝑓 𝑥𝑦 and 𝑓 𝑦𝑥 indicate that the original function has been partially differentiated with respect to one of the independent variables and that first partial derivative has in turn been partially differentiated with respect to the other independent variable.
This is denoted by 𝑓 𝑥𝑦 = 𝜕 𝜕𝑦 𝜕𝑧 𝜕𝑥 = 𝜕 2 𝑧 𝜕𝑥𝜕𝑦 and 𝑓 𝑦𝑥 = 𝜕 𝜕𝑥 𝜕𝑧 𝜕𝑦 = 𝜕 2 𝑧 𝜕𝑥𝜕𝑦
𝑓 𝑥𝑦 measures the rate of change of the first-order partial derivative 𝑓 𝑥 with respect to y while x is held constant
𝑓 𝑦𝑥 measures the rate of change of the first –order partial derivative 𝑓 𝑦 with respect to x while y is held constant
Note that 𝑓 𝑥𝑦 = 𝑓 𝑦𝑥 according to Young’s Theorem.

13. Examples
Find the first, second and cross partial derivatives for z= 7 𝑥 3 +9xy+2 𝑦 2
1. First order Partial Derivative:
𝑧 𝑥 = 21 𝑥 2 +9y and 𝑧 𝑦 =9x+4y
2. Second (direct) Partial Derivative:
𝑧 𝑥𝑥 =42x and 𝑧 𝑦𝑦 =4
3. Cross Partial Derivatives
𝑧 𝑥𝑦 = 𝜕 𝜕𝑦 𝜕 𝜕𝑥 = 𝜕 𝜕𝑦 (21 𝑥 2 +9y)=9
𝑧 𝑦𝑥 = 𝜕 𝜕𝑥 𝜕 𝜕𝑦 = 𝜕 𝜕𝑥 (9x+4y)=9

14. Total differentiation of multivariate functions

15. Definitions
For a function of two variables, the total differentiation measures the change in the dependent variable as a result of a small change in each of the independent variables.
If z= f(x,y), the total differential of z, denoted by dz is expressed mathematically as :
dz= 𝑧 𝑥 dx+ 𝑧 𝑦 dy
Where 𝑧 𝑥 and 𝑧 𝑦 are the partial derivatives with respect to xy and y respectively and dx and dy are small changes in x and y .
The total differential can thus be found by taking the partial derivatives of the function with respect to each independent variable and substituting in the formula stated above
To obtain the total derivative with respect to x, divide the formula through by dx to obtain: 𝑑𝑧 𝑑𝑥 = 𝑧 𝑥 𝑑𝑥 𝑑𝑥 + 𝑧 𝑦 𝑑𝑦 𝑑𝑥 = 𝑧 𝑥 + 𝑧 𝑦 𝑑𝑦 𝑑𝑥

16. Examples 1.Find the total differential of z= 𝑥 4 +8xy+3 𝑦 3
From the given function, 𝑧 𝑥 = 4 𝑥 3 +8y and 𝑧 𝑦 =8x+9 𝑦 2
Substituting in the formula: dz= 𝑧 𝑥 dx+ 𝑧 𝑦 dy, we have
dz= (4 𝑥 3 +8y )dx + (8x+9 𝑦 2 )dy
2. Find the total derivative 𝑑𝑧 𝑑𝑥 given z=f(x,y)=6 𝑥 3 +7y where y= g(x)=4 𝑥 2 +3x+8
Dividing the total derivative formula through by dx we have
𝑑𝑧 𝑑𝑥 = 𝑧 𝑥 + 𝑧 𝑦 𝑑𝑦 𝑑𝑥
From the question, 𝑧 𝑥 = 18 𝑥 2 and 𝑧 𝑦 = 7 and 𝑑𝑦 𝑑𝑥 = 8x+3
Substituting, we have:
𝑑𝑧 𝑑𝑥 = 18 𝑥 2 +7( 8x+3)= 18 𝑥 2 +56x+21

17. Unconstrained optimization

18. Unconstrained Optimization
Unconstrained Optimization involves finding the optimum to some decision problem in which there are no constraints.
For a multivariate function such as z=f(x,y) to have a minimum or maximum, the following three conditions have to be satisfied:
The first order partial derivatives must equal zero simultaneously. This ensures that at a given point (a,b) called the critical point, the function is neither increasing nor decreasing.
The second order direct partial derivatives, when evaluated at the critical point (a,b) must both be negative for a maximum and positive for a minimum. This ensures that at the critical point, the function is concave and moving downward in the case of a maximum and convex and moving upward in the case of a minimum.
The product of the second-order direct partial derivatives evaluated at the critical point must exceed the product of the cross partial derivatives also evaluated at the critical point.

19. Unconstrained Optimization: Summary of Conditions
For a Maximum:
𝑓 𝑥 , 𝑓 𝑦 =0
𝑓 𝑥𝑥 , 𝑓 𝑦𝑦 <0
𝑓 𝑥𝑥 . 𝑓 𝑦𝑦 > ( 𝑓 𝑥𝑦 ) Remember: 𝑓 𝑥𝑦 = 𝑓 𝑦𝑥
For a Minimum
𝑓 𝑥𝑥 , 𝑓 𝑦𝑦 >0
𝑓 𝑥𝑥 . 𝑓 𝑦𝑦 > ( 𝑓 𝑥𝑦 ) 2

20. Examples Optimize the function z=2 𝑦 3 - 𝑥 3 +147x-54y+12 by
finding the critical points, and testing whether the function is at its minimum or maximum.
For critical point, take the first order partial derivative, set them to zero and solve for x and y.
𝑧 𝑥 =-3 𝑥 = 𝑧 𝑦 =6 𝑦 2 -54=0
𝑥 2 = 𝑦 2 =9
x=± y=±3
There are four distinct critical points: (7,3), (7,-3),(-7,3) and (-7,-3)

21. Examples a. 𝑧 𝑥𝑥 7,3 =−42<0 𝑧 𝑦𝑦 (7,3)=36>0
Now take the second order direct partials and evaluate each of them at the critical points and check the signs.
𝑧 𝑥𝑥 =-6x 𝑧 𝑦𝑦 =12y
a. 𝑧 𝑥𝑥 7,3 =−42< 𝑧 𝑦𝑦 (7,3)=36>0
b. 𝑧 𝑥𝑥 7,−3 =−42< 𝑧 𝑦𝑦 (7,-3)=-36<0
c. 𝑧 𝑥𝑥 −7,3 =42> 𝑧 𝑦𝑦 (-7,3)=36>0
d. 𝑧 𝑥𝑥 −7,−3 =42> 𝑧 𝑦𝑦 (-7,-3)=-36<0
Diagnosis: a) and d) have different signs for the second order partial differential. This means that the second condition has not been satisfied which also hints that the third condition cannot be satisfied.
Therefore, the points (7,3) and (-7,-3) are neither maximum or minimum points. They are called saddle points.

22. Examples
The signs of the second direct partials are negative in b) and positive in c) which indicates that the function may be at a maximum at (7,-3) and a minimum at (-7,3)
We now test the third condition to be sure that this really holds true.
At point (7,-3): 𝑧 𝑥𝑥 7,−3 . 𝑧 𝑦𝑦 (7,-3) > ( 𝑧 𝑥𝑦 (7,−3)) 2
(-42). (-36)> 0 2
At point (-7,3): 𝑧 𝑥𝑥 −7,3 . 𝑧 𝑦𝑦 (7,-3) > ( 𝑧 𝑥𝑦 (−7,3)) 2
(42). (36)> 0 2
This means that the function is maximized at (7,-3) and minimized at (- 7,3)

23. Example: Economic Application
Sales (S) are a function of advertising in newspapers and magazines ( X, Y)
Max S = 200X + 100Y -10X2 -20Y2 +20XY
Solution:
Differentiate with respect to X and Y and set equal to zero.
¶S/¶X = X + 20Y=0
¶S/¶Y = Y + 20X = 0
solve for X & Y and Sales

24. Example Contd. S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250
Adding them, the -20X and +20X cancel, so we get Y = 0 → Y =15
Plug Y= 15 into (1):
X = 0, hence X = 25
To find Sales, plug X=25 and Y=15 into equation:
S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250

25. Trial Questions For each of the multivariable functions find
a. the critical points at which the function may be optimized
b. determine the nature of the critical points:
Z=3 𝑥 2 -xy+2 𝑦 2 -4x-7y+12
Z=60x+34y-4xy-6 𝑥 2 -3 𝑦 2 +5

26. Constrained optimization using lagrange multiplier

27. Constrained Optimization
In the optimization process, the variables to be chosen may be required to satisfy certain constraints.
For example, the quantities of goods demanded by a consumer may be subject to the available income level or his budget.
The lagrange muItiplier method is used to solve optimization problems involving constraints.
A constrained optimization problem is a problem of the form maximize (or minimize) the function f(x, y) subject to the condition g(x, y) = k ( a constant).
Since the constraint involves an equal sign, the optimization problem is described as equality-constrained optimization.

28. Constrained Optimization
In a constrained optimization problem, a new function F can be formed by setting the constraint to zero, multiplying it by λ( called the lagrange multiplier) and then adding the product to the original function : L(x,y, λ )= f(x,y)- λ[g(x,y)-c]
L(x,y, λ) is the Lagrangian function
f(x,y) is the objective function
g(x,y) is the constraint.

29. The Lagrange Multiplier
There are other techniques of solving constrained optimization problems.
However, this method is often used by economists because the lagrange multipliers have an important economic interpretation.
The lagrange multiplier approximates the marginal impact on the objective function caused by a small change in the constant of the constraint.
For example, if λ=2, it implies that a 1-unit increase/decrease in constant of the constraint would cause the objective function to increase/decrease by 2 units.
In utility maximization subject to a budget constraint for example, λ will estimate the marginal utility of an extra dollar of income.
Lagrange multipliers are also referred to as shadow prices.

30. The Lagrange Multiplier Method
To maximize or minimize f(x,y) subject to g(x,y)=k,
Follow the steps below:
Write down the Lagrangian Function:
L(x,y)= f(x,y)- λ(g(x,y)-k) λ is a constant
2. Differentiate L w.r.t x, y and λ. and equate the partial derivatives to 0. These are called the first order conditions.
3. The equations from the first order conditions above and the constraint yield the following three equations:
𝐿 1 ′ (x,y)= 𝑓 1 ′ (x,y)- λ 𝑔 1 ′ (x,y) =0
𝐿 2 ′ (x,y)= 𝑓 2 ′ (x,y)- λ 𝑔 2 ′ (x,y) =0
g(x,y)=k
4. Now solve these equations simultaneously for the three unknowns.

31. Example1: Constrained Optimization
Optimize the function Z= 4 𝑥 2 +3xy+6 𝑦 2 subject to the constraint
x +y =56.
Solution:
Lagrangian Function:
L(x,y, λ) = 4 𝑥 2 +3xy+6 𝑦 2 - λ(x+y-56)
First order conditions:
𝐿 𝑥 (x,y, λ)= 8x+3y- λ= (1)
𝐿 𝑦 (x,y, λ)=3x+12y- λ= (2)
x+y= (3)

32. Example1: Constrained Optimization
Subtract (2) from (1) to get: 5x-9y=0 => x=1.8y
substitute x=1.8y into (3) to get y=20.
This implies: x=36 and y=20 and λ= 348.
Substituting the values of x,y into the objective function gives z=

33. Example 2:
Use the Lagrangian multiplier to optimize the following function subject to the given constraint, and estimate the effect on the value of the objective function from a 1-unit change in the constant of the constraint.
Z=4 𝑥 2 -2xy+ 6𝑦 2 subject to x+y=72
Solution:
L(x,y, λ)=4 𝑥 2 -2xy+ 6𝑦 2 - λ(x+y-72)
First order conditions:
𝐿 𝑥 = 8x-2y- λ= (1)
𝐿 𝑦 = -2x+12y- λ= (2)
x+y= (3)

34. Example 2: Solving for x and y:
Subtract (2) from (1) to obtain: 10x-14y=0
x= 1.4y (4)
Now put (4) into (3)
1.4y+y=72
y=30
x=1.4(30)=42
The critical values are therefore: x=42 and y=30.
Substitute the critical values into (1) : 8(42)-2(30)=λ=276
Also substituting the critical values into the objective function yields: z=9936.

35. Example 2: Interpreting λ:
With λ=276, a 1-unit increase in the constant of the constraint will lead to an increase of approximately 276 in the value of the objective function.
The objective function will therefore increase in value from to 10,212.

36. Trial Questions Try These…
Use the Lagrangian multiplier to optimize the following functions subject to the given constraint, and estimate the effect on the value of the objective function from a 1-unit change in the constant of the constraint.
1. 26x-3 𝑥 2 +5xy- 6𝑦 2 +12y subject to 3x+y=170
2. 4xy 𝑧 2 subject to x+y+z=56
3. 4 𝑥 2 -3x+5xy-8y+2 𝑦 2 subject to x=2y

37. Constrained optimization of economic functions

38. Optimizing Economic Functions
Solutions to economic problems frequently have to be found under constraints. Examples include:
Maximizing utility subject to a budget constraint
Minimizing costs subject to some minimal requirement of output et.
The use of the Lagrangian function greatly facilitates the process of optimizing such objective functions in the face of their respective constraints.

39. Example 1: Cost Minimization
Find the critical values for minimizing the costs of a firm producing two goods x and y when the total cost function is given as C= 8 𝑥 2 -xy+12 𝑦 2 and the firm is bound by contract to produce a minimum combination of goods totaling 42 that is , subject to the constraint x+y=42.
Solution:
Setting the Lagrangian function:
L(x,y,λ)= 8 𝑥 2 -xy+12 𝑦 2 -λ(x+y-42)
First order conditions:
𝐿 𝑥 = 16x-y- λ=0
𝐿 𝑦 = -x+24y- λ=0
x+y=42

40. Example 1
Solving the three system of equations simultaneously, we obtain the following values: x=25, y=17 and λ=383. Interpretation of λ: a 1-unit increase in the constraint or production quota will lead to an increase in the cost of approximately $383. * You can check the nature of the critical values by using the second order conditions for a minimization from the previous sessions.

41. Example 2: Profit Maximization
A monopolistic firm has the following demand functions for each of its products x and y given as: x= 𝑃 𝑥 and y= 120- 𝑃 𝑦 . The combined cost function is c= 𝑥 2 +xy+ 𝑦 and the maximum joint production is 40. Therefore the production constraint is x+y=40. Find the profit-maximizing level of output, price and profit. Solution: x= 𝑃 𝑥 => 𝑃 𝑥 = 144-2x Y=120- 𝑃 𝑦 ⇒ 𝑃 𝑦 = 120-y Profit (𝜋)= TR-TC where TR=price x quantity => 𝑃 𝑥 . 𝑥+ 𝑃 𝑦 . 𝑦=(144-2x)x+(120-y)y (𝜋)= (144-2x)x+(120-y)y- ( 𝑥 2 +xy+ 𝑦 2 +35)

42. Example 2
The Lagrange Function: L(x,y,λ)= 144x-3 𝑥 2 -xy-2 𝑦 y-35- λ(x+y-40) First order conditions: 𝐿 𝑥 = 144-6x-y- λ=0 𝐿 𝑦 = -x-4y+120- λ=0 x+y=40 Solving the system of equations simultaneously yields: x=18, y=22 and λ=14 Substituting x and y into 𝑃 𝑥 and 𝑃 𝑦 gives: 𝑃 𝑥 =108 and 𝑃 𝑦 =98 and 𝜋=2861

43. Example 3: Utility Maximization
Maximize utility u= 𝑞 1 𝑞 2 when 𝑝 1 =1 and 𝑝 2 =4 and one’s budget B is Estimate a 1-unit increase in the budget.
Solution: The constraint can be written as:
𝑝 1 𝑞 1 + 𝑝 2 𝑞 2 = B
Substituting for the values above, the budget constraint is: 𝑞 1 +4 𝑞 2 = 120
The Lagrange function:
L( 𝑞 1 , 𝑞 2 ,λ)= 𝑞 1 𝑞 2 - λ( 𝑞 1 +4 𝑞 )

44. Example 3 Contd. First Order Conditions:
𝐿 𝑞 1 = 𝑞 2 - λ= => 𝑞 2 = λ
𝐿 𝑞 2 = 𝑞 1 - 4λ=0
𝑞 1 +4 𝑞 2 = 120
Solving the system of equations simultaneously yields:
𝑞 1 = 𝑞 2 = and λ=15
With λ=15, a unit increase of the budget constraint from 120 to 121 will lead to an increase in the utility function by approximately 15.

45. Example 4: Cobb Douglas Functions
Economic analysis usually makes use of the Cobb-Douglas production function written as : q= A 𝐾 𝛼 𝐿 𝛽
where A>0, 0<𝛼,𝛽<1.
where q is the quantity of output in physical units, K is the quantity of capital and L is the quantity of labour.
𝛼, the output elasticity of capital measures the percentage change in output as a result of a percentage change in K while L is held constant.
𝛽, the output elasticity of labour measures the percentage change in output as a result of a percentage change in labour while K is held constant.
A is and efficiency parameter which reflects the level of technology being applied in the production process.

46. Examples Optimize the Cobb-Douglas production function q= 𝐾 0.3 𝐿 0.5
subject to 6K+2L=384.
Solution:
The Lagrangian Function:
L(K,L,λ)= 𝐾 0.3 𝐿 λ(6K+2L-384)
First order conditions: hint: apply the power rule of differentiation
𝐿 𝐾 = 0.3 𝐾 −0.7 𝐿 λ= (1)
𝐿 𝐿 = 0.5 𝐾 0.3 𝐿 − λ= (2)
6K+2L= (3)

47. Examples Rearrange and divide (1) by (2) to eliminate λ:
0.3 𝐾 −0.7 𝐿 𝐾 0.3 𝐿 −0.5 = 6λ 2λ
0.6 𝐾 −1 𝐿 1 =3 => 𝐿 𝐾 = => L=5K
Now substituting L=5K into (3)
6K+2L=384 => 6K+2(5K)= 384 => 6K+10K=384
16K=384 => K=24 and L=5(24)=120.
K=24 and L=120

48. Example
A consumer has $600 to spend on two commodities, the first of which costs $20 per unit and the second $30 per unit. Suppose that the utility derived by the consumer from x units of the first commodity and y units of the second commodity is given by the Cobb-Douglas utility function U(x, y) = 10 𝑥 0.6 𝑦 How many units of each commodity should the consumer buy to maximize utility? Optimal values of x and y: X=18 and y=8

49. Trial Questions
Maximize the following utility functions subject to the given budget constraints.
1. u= 𝑥 0.6 𝑦 given 𝑝 𝑥 =8, 𝑝 𝑦 =5 and B=680
2. u= 𝑥 0.8 𝑦 given 𝑝 𝑥 =5, 𝑝 𝑦 =3 and B=75