Sure, let's solve the given problem step-by-step.
### (a) Finding the amount of money the company should spend on advertising to yield maximum profit:
1. Profit Function:
The profit [tex]\( P \)[/tex] given as a function of advertising expenditure [tex]\( s \)[/tex] is:
[tex]\[
P = -\frac{1}{10}s^3 + 12s^2 + 450
\][/tex]
2. First Derivative (Finding Critical Points):
To find the critical points where profit could be at a maximum or minimum, we first find the first derivative of [tex]\( P \)[/tex] with respect to [tex]\( s \)[/tex]:
[tex]\[
\frac{dP}{ds} = -\frac{3}{10}s^2 + 24s
\][/tex]
3. Setting the First Derivative to Zero:
Set the first derivative equal to zero to find the critical points:
[tex]\[
-\frac{3}{10}s^2 + 24s = 0
\][/tex]
Solve for [tex]\( s \)[/tex]:
[tex]\[
s(-\frac{3}{10}s + 24) = 0
\][/tex]
This gives us:
[tex]\[
s = 0 \quad \text{or} \quad s = 80
\][/tex]
4. Second Derivative (Determining Concavity):
Next, we find the second derivative of [tex]\( P \)[/tex] to determine concavity and thus confirm if the critical point [tex]\( s = 80 \)[/tex] is a maximum:
[tex]\[
\frac{d^2P}{ds^2} = -\frac{6}{10}s + 24 = -0.6s + 24
\][/tex]
5. Evaluating Second Derivative at Critical Points:
- At [tex]\( s = 0 \)[/tex]:
[tex]\[
\frac{d^2P}{ds^2} = -0.6 \cdot 0 + 24 = 24 \quad (\text{positive concavity})
\][/tex]
- At [tex]\( s = 80 \)[/tex]:
[tex]\[
\frac{d^2P}{ds^2} = -0.6 \cdot 80 + 24 = -48 + 24 = -24 \quad (\text{negative concavity})
\][/tex]
Since the second derivative is negative at [tex]\( s = 80 \)[/tex], this means [tex]\( s = 80 \)[/tex] is a local maximum.
### Conclusion for (a):
The company should spend \$80,000 on advertising to yield a maximum profit.
[tex]\[
\boxed{80,000}
\][/tex]
### (b) Finding the point of diminishing returns:
1. Point of Diminishing Returns:
This is where the second derivative changes sign, usually where the second derivative is zero.
[tex]\[
-0.6s + 24 = 0
\][/tex]
2. Solving for [tex]\( s \)[/tex]:
[tex]\[
s = \frac{24}{0.6} = 40
\][/tex]
3. Calculating [tex]\( \rho \)[/tex] at [tex]\( s = 40 \)[/tex]:
[tex]\[
P = -\frac{1}{10}(40)^3 + 12(40)^2 + 450
\][/tex]
[tex]\[
P = -\frac{1}{10}(64000) + 19200 + 450
\][/tex]
[tex]\[
P = -6400 + 19200 + 450 = 13250
\][/tex]
### Conclusion for (b):
The point of diminishing returns is when [tex]\( s = 40,000 \)[/tex] dollars, and the corresponding profit [tex]\( \rho \)[/tex] is [tex]\( 13,250,000 \)[/tex] dollars.
[tex]\[
(s, \rho) = \boxed{(40, 13250)}
\][/tex]
