Answer :
Sure, let's approach each part of the problem step-by-step.
### Part (a): Finding the Revenue Function
The marginal revenue (MR) function is given by:
[tex]\[ \operatorname{MR}(x) = 4x \left(x^2 + 28,000\right)^{-\frac{2}{3}} \][/tex]
The revenue function, [tex]\( R(x) \)[/tex], is the integral of the marginal revenue function:
[tex]\[ R(x) = \int \operatorname{MR}(x) \, dx \][/tex]
So, we need to integrate [tex]\( \operatorname{MR}(x) \)[/tex]:
[tex]\[ R(x) = \int 4x \left(x^2 + 28,000\right)^{-\frac{2}{3}} \, dx \][/tex]
To integrate this, we use the substitution method. Let:
[tex]\[ u = x^2 + 28,000 \][/tex]
Then:
[tex]\[ du = 2x \, dx \quad \Rightarrow \quad \frac{du}{2} = x \, dx \][/tex]
Substituting [tex]\( u \)[/tex] and [tex]\( \frac{du}{2} \)[/tex] into the integral, we get:
[tex]\[ R(x) = \int 4x \left(u\right)^{-\frac{2}{3}} \cdot \frac{du}{2x} = 2 \int u^{-\frac{2}{3}} \, du \][/tex]
Now, integrate [tex]\( u^{-\frac{2}{3}} \)[/tex]:
[tex]\[ \int u^{-\frac{2}{3}} \, du = u^{1 - \frac{2}{3}} \cdot \frac{3}{1} = 3u^{\frac{1}{3}} \][/tex]
So:
[tex]\[ R(x) = 2 \cdot 3u^{\frac{1}{3}} + C = 6u^{\frac{1}{3}} + C \][/tex]
Substitute back [tex]\( u = x^2 + 28,000 \)[/tex]:
[tex]\[ R(x) = 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} + C \][/tex]
We need to find [tex]\( C \)[/tex]. We know [tex]\( R(130) = 43,255 \)[/tex] thousands of dollars.
Substitute [tex]\( x = 130 \)[/tex] and [tex]\( R(130) \)[/tex]:
[tex]\[ 43,255 = 6 \left(130^2 + 28,000\right)^{\frac{1}{3}} + C \][/tex]
Calculate [tex]\( 130^2 + 28,000 \)[/tex]:
[tex]\[ 130^2 = 16,900 \quad \Rightarrow \quad 130^2 + 28,000 = 44,900 \][/tex]
Thus:
[tex]\[ 43,255 = 6 \left(44,900\right)^{\frac{1}{3}} + C \][/tex]
Compute [tex]\( (44,900)^{\frac{1}{3}} \)[/tex]:
[tex]\[ (44,900)^{\frac{1}{3}} \approx 35.796 \][/tex]
So:
[tex]\[ 43,255 = 6 \cdot 35.796 + C \][/tex]
[tex]\[ 43,255 \approx 214.776 + C \][/tex]
[tex]\[ C \approx 43,255 - 214.776 \approx 43,040.224 \][/tex]
Therefore, the revenue function is:
[tex]\[ R(x) \approx 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} + 43,040.224 \][/tex]
### Part (b): Revenue from Selling 250 Gadgets
Now, we need to find [tex]\( R(250) \)[/tex]:
[tex]\[ R(250) = 6 \left(250^2 + 28,000\right)^{\frac{1}{3}} + 43,040.224 \][/tex]
Compute [tex]\( 250^2 + 28,000 \)[/tex]:
[tex]\[ 250^2 = 62,500 \quad \Rightarrow \quad 250^2 + 28,000 = 90,500 \][/tex]
Now:
[tex]\[ (90,500)^{\frac{1}{3}} \approx 44.721 \][/tex]
So:
[tex]\[ R(250) \approx 6 \cdot 44.721 + 43,040.224 \][/tex]
[tex]\[ R(250) \approx 268.326 + 43,040.224 \approx 43,308.55 \text{ thousands of dollars} \][/tex]
### Part (c): Gadgets Sold for at Least [tex]$45,000 We need to solve for \( x \) in: \[ R(x) = 45,000 \quad \Rightarrow \quad 45,000 = 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} + 43,040.224 \] Rearrange to solve for \( \left(x^2 + 28,000\right)^{\frac{1}{3}} \): \[ 45,000 - 43,040.224 = 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} \] \[ 1,959.776 = 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} \] \[ 326.629 \approx \left(x^2 + 28,000\right)^{\frac{1}{3}} \] Now, cube both sides: \[ 326.629^3 \approx x^2 + 28,000 \] Calculate \( 326.629^3 \): \[ 326.629^3 \approx 34,867,467 \] Thus: \[ 34,867,467 - 28,000 = x^2 \] \[ x^2 \approx 34,839,467 \] \[ x \approx \sqrt{34,839,467} \approx 5,904 \] So, to achieve a revenue of at least \$[/tex]45,000, approximately 5,904 gadgets must be sold.
### Part (a): Finding the Revenue Function
The marginal revenue (MR) function is given by:
[tex]\[ \operatorname{MR}(x) = 4x \left(x^2 + 28,000\right)^{-\frac{2}{3}} \][/tex]
The revenue function, [tex]\( R(x) \)[/tex], is the integral of the marginal revenue function:
[tex]\[ R(x) = \int \operatorname{MR}(x) \, dx \][/tex]
So, we need to integrate [tex]\( \operatorname{MR}(x) \)[/tex]:
[tex]\[ R(x) = \int 4x \left(x^2 + 28,000\right)^{-\frac{2}{3}} \, dx \][/tex]
To integrate this, we use the substitution method. Let:
[tex]\[ u = x^2 + 28,000 \][/tex]
Then:
[tex]\[ du = 2x \, dx \quad \Rightarrow \quad \frac{du}{2} = x \, dx \][/tex]
Substituting [tex]\( u \)[/tex] and [tex]\( \frac{du}{2} \)[/tex] into the integral, we get:
[tex]\[ R(x) = \int 4x \left(u\right)^{-\frac{2}{3}} \cdot \frac{du}{2x} = 2 \int u^{-\frac{2}{3}} \, du \][/tex]
Now, integrate [tex]\( u^{-\frac{2}{3}} \)[/tex]:
[tex]\[ \int u^{-\frac{2}{3}} \, du = u^{1 - \frac{2}{3}} \cdot \frac{3}{1} = 3u^{\frac{1}{3}} \][/tex]
So:
[tex]\[ R(x) = 2 \cdot 3u^{\frac{1}{3}} + C = 6u^{\frac{1}{3}} + C \][/tex]
Substitute back [tex]\( u = x^2 + 28,000 \)[/tex]:
[tex]\[ R(x) = 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} + C \][/tex]
We need to find [tex]\( C \)[/tex]. We know [tex]\( R(130) = 43,255 \)[/tex] thousands of dollars.
Substitute [tex]\( x = 130 \)[/tex] and [tex]\( R(130) \)[/tex]:
[tex]\[ 43,255 = 6 \left(130^2 + 28,000\right)^{\frac{1}{3}} + C \][/tex]
Calculate [tex]\( 130^2 + 28,000 \)[/tex]:
[tex]\[ 130^2 = 16,900 \quad \Rightarrow \quad 130^2 + 28,000 = 44,900 \][/tex]
Thus:
[tex]\[ 43,255 = 6 \left(44,900\right)^{\frac{1}{3}} + C \][/tex]
Compute [tex]\( (44,900)^{\frac{1}{3}} \)[/tex]:
[tex]\[ (44,900)^{\frac{1}{3}} \approx 35.796 \][/tex]
So:
[tex]\[ 43,255 = 6 \cdot 35.796 + C \][/tex]
[tex]\[ 43,255 \approx 214.776 + C \][/tex]
[tex]\[ C \approx 43,255 - 214.776 \approx 43,040.224 \][/tex]
Therefore, the revenue function is:
[tex]\[ R(x) \approx 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} + 43,040.224 \][/tex]
### Part (b): Revenue from Selling 250 Gadgets
Now, we need to find [tex]\( R(250) \)[/tex]:
[tex]\[ R(250) = 6 \left(250^2 + 28,000\right)^{\frac{1}{3}} + 43,040.224 \][/tex]
Compute [tex]\( 250^2 + 28,000 \)[/tex]:
[tex]\[ 250^2 = 62,500 \quad \Rightarrow \quad 250^2 + 28,000 = 90,500 \][/tex]
Now:
[tex]\[ (90,500)^{\frac{1}{3}} \approx 44.721 \][/tex]
So:
[tex]\[ R(250) \approx 6 \cdot 44.721 + 43,040.224 \][/tex]
[tex]\[ R(250) \approx 268.326 + 43,040.224 \approx 43,308.55 \text{ thousands of dollars} \][/tex]
### Part (c): Gadgets Sold for at Least [tex]$45,000 We need to solve for \( x \) in: \[ R(x) = 45,000 \quad \Rightarrow \quad 45,000 = 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} + 43,040.224 \] Rearrange to solve for \( \left(x^2 + 28,000\right)^{\frac{1}{3}} \): \[ 45,000 - 43,040.224 = 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} \] \[ 1,959.776 = 6 \left(x^2 + 28,000\right)^{\frac{1}{3}} \] \[ 326.629 \approx \left(x^2 + 28,000\right)^{\frac{1}{3}} \] Now, cube both sides: \[ 326.629^3 \approx x^2 + 28,000 \] Calculate \( 326.629^3 \): \[ 326.629^3 \approx 34,867,467 \] Thus: \[ 34,867,467 - 28,000 = x^2 \] \[ x^2 \approx 34,839,467 \] \[ x \approx \sqrt{34,839,467} \approx 5,904 \] So, to achieve a revenue of at least \$[/tex]45,000, approximately 5,904 gadgets must be sold.
