Interpretations of the Derivative

[AInerd72]

QUESTION TEXT: A company's revenue from car sales, C(in thousands of dollars), is a function of advertising expenditure, a, in thousands of dollars, so C = f(a).

a) What does the company hope is true about the sign of f'?

b) What does the statement f'(100) = 2 mean in practical terms? How about f'(100) = 0.5

 

[AInerd72]

QUESTION TEXT: A company's revenue from car sales, C(in thousands of dollars), is a function of advertising expenditure, a, in thousands of dollars, so C = f(a).

a) What does the company hope is true about the sign of f'?

b) What does the statement f'(100) = 2 mean in practical terms? How about f'(100) = 0.5

c) Suppose the company plans to spend about $100,000 on advertising. If f'(100) = 2, should the company spend more or less then $100,000 on advertising? What if f'(100) = 0.5?

MY (PARTIALLY CORRECT) ATTEMPT: Alright, so part (a) completely confuses me. At first I thought zero, since that would represent an optimum. However, the book solution is "always positive" with no explanation. Why would this be the case?

For part (b) I thought f'(100) = 2 meant revenue would increase 2 thousand dollars by spending 101 thousand dollars (an additional thousand), and f'(100) = 0.5 meant revenue would increase five hundred dollars by spending 101 thousand dollars (an additional thousand). [CORRECT]

For part (c) I knew that the company should spend more in the first case and less in the second case. [CORRECT]

In short, it is only part (a) that confuses me because why would it always be positive, wouldn't that mean that they ought to be spending more to move towards an optimum?

 

[Ishuda]

QUESTION TEXT: A company's revenue from car sales, C(in thousands of dollars), is a function of advertising expenditure, a, in thousands of dollars, so C = f(a).

a) What does the company hope is true about the sign of f'?

b) What does the statement f'(100) = 2 mean in practical terms? How about f'(100) = 0.5

c) Suppose the company plans to spend about $100,000 on advertising. If f'(100) = 2, should the company spend more or less then $100,000 on advertising? What if f'(100) = 0.5?

MY (PARTIALLY CORRECT) ATTEMPT: Alright, so part (a) completely confuses me. At first I thought zero, since that would represent an optimum. However, the book solution is "always positive" with no explanation. Why would this be the case?

For part (b) I thought f'(100) = 2 meant revenue would increase 2 thousand dollars by spending 101 thousand dollars (an additional thousand), and f'(100) = 0.5 meant revenue would increase five hundred dollars by spending 101 thousand dollars (an additional thousand). [CORRECT]

For part (c) I knew that the company should spend more in the first case and less in the second case. [CORRECT]

In short, it is only part (a) that confuses me because why would it always be positive, wouldn't that mean that they ought to be spending more to move towards an optimum?

Think about your answer to part(b). Basically you said 'if they spend more, they get more than they spend'. So put that in a formula: Spend an additional dx, then revenue at x+dx, i.e. f(x+dx), should be larger than that at x, i.e larger than f(x), plus dx
f(x+dx) > f(x) + dx
or
$\displaystyle \frac{f(x+dx) - f(x)}{dx}\, \gt\, 0$

So, what does that say about the derivative of f?