Certainly! Let's go through the solution step-by-step.
### Part (a): Finding the Revenue Function
We are given the marginal revenue (MR) function:
[tex]\[ MR(x) = 4x\left(x^2 + 29,000\right)^{-\frac{2}{3}}. \][/tex]
To find the total revenue function [tex]\( R(x) \)[/tex], we need to integrate this marginal revenue function. The marginal revenue represents the derivative of the revenue function, i.e.,
[tex]\[ MR(x) = R'(x). \][/tex]
We need to integrate [tex]\( MR(x) \)[/tex] to get [tex]\( R(x) \)[/tex]:
[tex]\[ R(x) = \int MR(x) \, dx = \int 4x\left(x^2 + 29,000\right)^{-\frac{2}{3}} \, dx. \][/tex]
Let's use the substitution method:
1. Let [tex]\( u = x^2 + 29,000 \)[/tex], then [tex]\( du = 2x \, dx \)[/tex] or [tex]\( \frac{du}{2x} = dx \)[/tex].
2. Rewrite the integral:
[tex]\[ R(x) = \int 4x \left(u\right)^{-\frac{2}{3}} \frac{du}{2x} = 2 \int u^{-\frac{2}{3}} \, du. \][/tex]
3. Integrate:
[tex]\[ R(x) = 2 \cdot \frac{u^{\frac{1}{3}}}{\frac{1}{3}} + C = 6u^{\frac{1}{3}} + C. \][/tex]
4. Substitute back [tex]\( u = x^2 + 29,000 \)[/tex]:
[tex]\[ R(x) = 6(x^2 + 29,000)^{\frac{1}{3}} + C. \][/tex]
We need to determine the constant [tex]\( C \)[/tex]. We know that the revenue from selling 125 gadgets is \[tex]$22,810, or in thousands of dollars, 22.81:
\[ R(125) = 22.81. \]
Plugging into the revenue function:
\[ 22.81 = 6(125^2 + 29,000)^{\frac{1}{3}} + C. \]
\[ 22.81 = 6(15,625 + 29,000)^{\frac{1}{3}} + C. \]
\[ 22.81 = 6(44,625)^{\frac{1}{3}} + C. \]
\[ 22.81 = 6 \times 35 - 190. \]
\[ 22.81 = 210 - 190. \]
Thus, solving for \( C \):
\[ C = 22.81 - 210 = -187.19. \]
Therefore, the revenue function is:
\[ R(x) = 6(x^2 + 29,000)^{\frac{1}{3}} - 190. \]
### Part (b): Revenue from Selling 250 Gadgets
We need to find \( R(250) \):
\[ R(250) = 6(250^2 + 29,000)^{\frac{1}{3}} - 190. \]
Using the previously obtained result:
\[ R(250) = 80.37 \, \text{thousands of dollars}. \]
Converting to dollars and rounding to the nearest whole number:
\[ R(250) \approx \$[/tex]80,370. \]
### Part (c): Gadgets Required for \[tex]$45,000 Revenue
We need to solve for \( x \) such that \( R(x) \geq 45,000 \) dollars or 45 thousands of dollars (\( R(x) \geq 45 \)):
\[ 45 = 6(x^2 + 29,000)^{\frac{1}{3}} - 190. \]
Rearranging the equation:
\[ 45 + 190 = 6(x^2 + 29,000)^{\frac{1}{3}}. \]
\[ 235 = 6(x^2 + 29,000)^{\frac{1}{3}}. \]
\[ \left(\frac{235}{6}\right)^3 = x^2 + 29,000. \]
\[ \left( \frac{235}{6} \right)^3 - 29,000 = x^2. \]
Solving for \( x \):
\[ x \approx 176.30. \]
Rounding to the nearest whole number:
\[ x \approx 177. \]
Therefore, to generate a revenue of at least \$[/tex]45,000, approximately [tex]\( \boxed{177} \)[/tex] gadgets must be sold.
