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Case study for final exam, Assignments of Mathematics

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Typology: Assignments

2022/2023

Uploaded on 07/11/2024

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Case Study: Optimizing Advertising Spend for a Product
Problem Statement
A company sells a product and wants to maximize its profit by optimally allocating its
advertising budget between two channels: online ads (O) and TV ads (T). The profit (P)
generated from the product sales depends on the amount of money spent on these two
advertising channels.
Objective
To find the optimal amounts of spending on online ads (O) and TV ads (T) that maximize the
profit (P).
Profit Function
The profit function is given by: P(O,T)=20O0.5+30T0.5−O−T
Here:
20O0.5 represents the revenue generated from online ads.
30T0.5 represents the revenue generated from TV ads.
O and T are the expenditures on online and TV ads respectively.
The terms −O and −T represent the costs associated with these expenditures.
Solution Approach
To maximize the profit, we need to find the values of O and T that maximize the function
P(O,T). This involves finding the critical points of P(O,T) and determining whether these
points are maxima, minima, or saddle points.
Detailed Steps
1. Compute the Partial Derivatives:
Po=0 ∂P/∂O=0 10O-0.5 -1 =0 O = 100
PT=0 ∂P/∂T=0 15T-0.5−1 = 0 T = 225
So, the stationary point is (O,T)=(100,225)
2. Second Partial Derivative Test:
P’’OO ( 100,225) = -5*100-1.5= -0.05
P’’TT ( 100,225) = -7.5*225-1.5 = -0.01
P’’OT (100,225) = 0
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Case Study: Optimizing Advertising Spend for a Product

Problem Statement

A company sells a product and wants to maximize its profit by optimally allocating its advertising budget between two channels: online ads (O) and TV ads (T). The profit (P) generated from the product sales depends on the amount of money spent on these two advertising channels.

Objective

To find the optimal amounts of spending on online ads (O) and TV ads (T) that maximize the profit (P).

Profit Function

The profit function is given by: P(O,T)=20O0.5+30T0.5−O−T Here:  20O0.5^ represents the revenue generated from online ads.  30T0.5^ represents the revenue generated from TV ads.  O and T are the expenditures on online and TV ads respectively.  The terms −O and −T represent the costs associated with these expenditures.

Solution Approach

To maximize the profit, we need to find the values of O and T that maximize the function P(O,T). This involves finding the critical points of P(O,T) and determining whether these points are maxima, minima, or saddle points.

Detailed Steps

  1. Compute the Partial Derivatives: P’o=0 ∂P/∂O= 0 10O-0.5^ -1 =0  O = 100 P’T=0 ∂P/∂T= 0  15T-0.5−1 = 0  T = 225 So, the stationary point is (O,T)=(100,225)
  2. Second Partial Derivative Test: P’’OO ( 100,225) = -5100-1.5= -0. P’’TT ( 100,225) = -7.5225-1.5^ = -0. P’’OT (100,225) = 0

Δ = (-0.05)*(-0.01)-0^2 = 0.0005 > 0

Since Δ > 0 and A < 0, we conclude that the point (100,225) maximize the function.

Conclusion

The optimal amounts of spending on online ads and TV ads to maximize the profit are O= 100 and T=225. At these expenditures, the company achieves the maximum profit.

Key Points

Partial differentiation helps us understand how changes in each advertising channel affect the profit.  By analyzing the partial derivatives and using the second partial derivative test, we identify the optimal spending amounts that maximize profit.  This approach can be applied to various scenarios involving the optimization of resources to achieve a desired outcome. This case study demonstrates how partial differentiation is applied in unconstrained optimization to find and classify critical points for maximizing a profit function.