Question

The profit P (in thousands of dollars) for a company spending an amount s (in thousands of dollars on advertising is
P= -(1/10)s^3 + 6s^2 + 400
Find the amount of money the company should spend on advertising in order to yield a maximum profit. The point of diminishing returns is the point at which the rate of growth of the profit function begins to decline. Find the point of diminishing returns.

Answer 1

Answer:

The company should spend $40 to yield a maximum profit.

The point of diminishing returns is (40, 3600).

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right  

Equality Properties

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtraction Property of Equality

Algebra I

Coordinate Planes

• Coordinates (x, y) → (s, P)

Functions

• Function Notation

Terms/Coefficients

• Factoring/Expanding

Quadratics

Algebra II

Coordinate Planes

• Maximums/Minimums

Calculus

Derivatives

• Derivative Notation

Derivative Property [Multiplied Constant]:                                                           $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]:                                                         $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$  

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

1st Derivative Test - tells us where on the function f(x) does it have a relative maximum or minimum

• Critical Numbers

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle P = \frac{-1}{10}s^3 + 6s^2 + 400$

Step 2: Differentiate

1. [Function] Derivative Property [Addition/Subtraction]:                               $\displaystyle P' = \frac{dP}{ds} \bigg[ \frac{-1}{10}s^3 \bigg] + \frac{dP}{ds} [ 6s^2 ] + \frac{dP}{ds} [ 400 ]$
2. [Derivative] Rewrite [Derivative Property - Multiplied Constant]:               $\displaystyle P' = \frac{-1}{10} \frac{dP}{ds} \bigg[ s^3 \bigg] + 6 \frac{dP}{ds} [ s^2 ] + \frac{dP}{ds} [ 400 ]$
3. [Derivative] Basic Power Rule:                                                                     $\displaystyle P' = \frac{-1}{10}(3s^2) + 6(2s)$
4. [Derivative] Simplify:                                                                                     $\displaystyle P' = -\frac{3s^2}{10} + 12s$

Step 3: 1st Derivative Test

1. [Derivative] Set up:                                                                                       $\displaystyle 0 = -\frac{3s^2}{10} + 12s$
2. [Derivative] Factor:                                                                                       $\displaystyle 0 = \frac{-3s(s - 40)}{10}$
3. [Multiplication Property of Equality] Isolate s terms:                                   $\displaystyle 0 = -3s(s - 40)$
4. [Solve] Find quadratic roots:                                                                         $\displaystyle s = 0, 40$

∴ s = 0, 40 are our critical numbers.

Step 4: Find Profit

1. [Function] Substitute in s = 0:                                                                       $\displaystyle P(0) = \frac{-1}{10}(0)^3 + 6(0)^2 + 400$
2. [Order of Operations] Evaluate:                                                                   $\displaystyle P(0) = 400$
3. [Function] Substitute in s = 40:                                                                     $\displaystyle P(40) = \frac{-1}{10}(40)^3 + 6(40)^2 + 400$
4. [Order of Operations] Evaluate:                                                                   $\displaystyle P(40) = 3600$

We see that we will have a bigger profit when we spend s = $40.

∴ The maximum profit is $3600.

∴ The point of diminishing returns is ($40, $3600).

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation (Applications)