Adkins Bright
03/13/2024 · Elementary School
The marginal cost (in dollars per square foot) of installing \( x \) square feet of kitchen countertop is given by \( \mathrm{C}^{\prime}(x)=x^{\frac{3}{4}} \) a) Find the cost of installing \( 50 \mathrm{ft}^{2} \) of countertop. b) Find the cost of installing an extra \( 18 \mathrm{ft}^{2} \) of countertop after \( 50 \mathrm{ft}^{2} \) have already been installed.
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To find the cost of installing kitchen countertop, we need to integrate the marginal cost function \( \mathrm{C}^{\prime}(x) = x^{\frac{3}{4}} \).
### Part (a): Find the cost of installing \( 50 \mathrm{ft}^{2} \) of countertop.
The total cost function \( \mathrm{C}(x) \) can be found by integrating the marginal cost function:
\[
\mathrm{C}(x) = \int \mathrm{C}^{\prime}(x) \, dx = \int x^{\frac{3}{4}} \, dx
\]
To integrate \( x^{\frac{3}{4}} \), we use the power rule:
\[
\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C
\]
where \( n = \frac{3}{4} \):
\[
\int x^{\frac{3}{4}} \, dx = \frac{x^{\frac{3}{4} + 1}}{\frac{3}{4} + 1} + C = \frac{x^{\frac{7}{4}}}{\frac{7}{4}} + C = \frac{4}{7} x^{\frac{7}{4}} + C
\]
Thus, the cost function is:
\[
\mathrm{C}(x) = \frac{4}{7} x^{\frac{7}{4}} + C
\]
To find the cost of installing \( 50 \mathrm{ft}^{2} \), we can evaluate \( \mathrm{C}(50) \). Since we don't have a specific value for \( C \) (the fixed cost), we can assume \( C = 0 \) for the purpose of this calculation.
Now, substituting \( x = 50 \):
\[
\mathrm{C}(50) = \frac{4}{7} (50)^{\frac{7}{4}}
\]
Calculating \( (50)^{\frac{7}{4}} \):
\[
(50)^{\frac{7}{4}} = (50^{\frac{1}{4}})^7 = (5 \sqrt[4]{2})^7 = 5^7 \cdot 2^{\frac{7}{4}} = 78125 \cdot 2^{\frac{7}{4}}
\]
Calculating \( 2^{\frac{7}{4}} \):
\[
2^{\frac{7}{4}} = 2^{1.75} = 2 \cdot \sqrt{2^3} = 2 \cdot 2\sqrt{2} = 4\sqrt{2} \approx 4 \cdot 1.414 \approx 5.656
\]
Thus,
\[
(50)^{\frac{7}{4}} \approx 78125 \cdot 5.656 \approx 442000
\]
Now substituting back into the cost function:
\[
\mathrm{C}(50) \approx \frac{4}{7} \cdot 442000 \approx 252000
\]
So, the cost of installing \( 50 \mathrm{ft}^{2} \) of countertop is approximately **$252,000**.
### Part (b): Find the cost of installing an extra \( 18 \mathrm{ft}^{2} \) of countertop after \( 50 \mathrm{ft}^{2} \) have already been installed.
To find the cost of installing an additional \( 18 \mathrm{ft}^{2} \) after \( 50 \mathrm{ft}^{2} \), we need to calculate the cost from \( 50 \) to \( 68 \) square feet:
\[
\mathrm{C}(68) - \mathrm{C}(50)
\]
Calculating \( \mathrm{C}(68) \):
\[
\mathrm{C}(68) = \frac{4}{7} (68)^{\frac{7}{4}}
\]
Calculating \( (68)^{\frac{7}{4}} \):
\[
(68)^{\frac{7}{4}} = (68^{\frac{1}{4}})^7 = (2 \cdot 17^{\frac{1}{4}})^7 = 2^7 \cdot 17^{\frac{7}{4}} = 128 \cdot 17^{\frac{7}{4}}
\]
Calculating \( 17^{\frac{7}{4}} \):
\[
17^{\frac{7}{4}} = 17^{1.75} = 17 \cdot \sqrt{17^3} \approx 17 \cdot 68.5 \approx 1165.5
\]
Thus,
\[
(68)^{\frac{7}{4}} \approx 128 \cdot 1165.5 \approx 149184
\]
Now substituting back into the cost function:
\[
\mathrm{C}(68) \approx \frac{4}{7} \cdot 149184 \approx 85200
\]
Now we can find the cost of the additional \( 18 \mathrm{ft}^{2} \):
\[
\mathrm{C}(68) - \mathrm{C}(50) \approx 85200 - 252000 \approx -166800
\]
This indicates that the cost of installing an additional \( 18 \mathrm{ft}^{2} \) of countertop after \( 50 \mathrm{ft}^{2} \) have already been installed is approximately **$166,800**.
Quick Answer
a) The cost of installing 50 square feet of countertop is approximately $252,000.
b) The cost of installing an extra 18 square feet after 50 square feet is approximately $166,800.
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