IDNLearn.com is the perfect place to get detailed and accurate answers to your questions. Get the information you need from our community of experts, who provide detailed and trustworthy answers.
Sagot :
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.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.
