MAT 142 Mathematics

Question:

An auto dealership keeps track of how much advertising it spends each month and the monthly revenue.

Below is a table that shows both the probable revenues and advertising expenditures.

Advertising

(millions of dollars)

Revenue

(thousands upon thousands of dollars)

Find a cubic Polynomial that models data.

One graph of the function should be provided.

You can round up to three significant numbers.

Avoid using y or x. Find the first derivative, and explain its meaning in terms of data.

Find the relative extremes of the problem and explain the meaning.

Use proper calculus notation. Find the point where diminishing returns occur and explain the meaning in terms of your problem.

Answer:

dRevenue

The rate of change in revenue per unit of Advertising. This is the change in dAdvertising when Advertising expenditures are reduced by one dollar.

If the first derivative is null, it means that advertising has achieved its highest or lowest revenue.

c) We can convert the first derivative of the equation to 0. This gives us: = –1.041Avertising2 + 24,184Advertsing — 9.953 = 0 Solving quadratic equation.

Extreme Advertising Values = 24184 – N W676 – 5.38 (Thousands of Dollars), 2082

or — 24184+.11W6 – 17.85(Thousdrid Dollars) 2082 Revenue = 193.85 (thousands of dollars), using the equation. On the other side, Advertsing = 17.85 or more, Revenue = 530.21%.

dRevenue = dAdvertising 2.082Advertising + 24.184

The second derivative indicates how much marginal revenue will change with an increase or decrease of advertising costs.

The local maximum is the point at which diminishing returns occur.

The point at which the return begins to diminish is when advertising costs exceed $17850 dollars and revenue exceeds $530210.

Therefore, any additional advertising costs incurred after $17850 dollars have been spent will not result in an increase or decrease in revenue.

Any advertising cost greater than $17850 leads to revenue below $530210.

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