The marginal cost (in dollars per square foot) of installing \( x \) square feet of kitchen countertop is given by \( \mathrm{C}^{\prime}(\mathrm{x})=\mathrm{x}^{\frac{5}{6}} \) a) Find the cost of installing \( 60 \mathrm{ft}^{2} \) of countertop. b) Find the cost of installing an extra \( 13 \mathrm{ft}^{2} \) of countertop after \( 60 \mathrm{ft}^{2} \) have already been installed. a) Set up the integral for the cost of installing \( 60 \mathrm{ft}^{2} \) of countertop. c( 60\( )=\int_{0} \square \mathrm{dx} \)

To find the cost of installing \( 60 \, \text{ft}^2 \) of kitchen countertop, we need to integrate the marginal cost function \( C'(x) = x^{\frac{5}{6}} \) from \( 0 \) to \( 60 \). ### a) Set up the integral for the cost of installing \( 60 \, \text{ft}^2 \) of countertop. The cost function \( C(x) \) can be found by integrating the marginal cost function: \[ C(60) = \int_{0}^{60} C'(x) \, dx = \int_{0}^{60} x^{\frac{5}{6}} \, dx \] ### b) Find the cost of installing \( 60 \, \text{ft}^2 \) of countertop. Now we will evaluate the integral: \[ C(60) = \int_{0}^{60} x^{\frac{5}{6}} \, dx \] To compute this integral, we first find the antiderivative of \( x^{\frac{5}{6}} \): \[ \int x^{\frac{5}{6}} \, dx = \frac{x^{\frac{5}{6} + 1}}{\frac{5}{6} + 1} = \frac{x^{\frac{11}{6}}}{\frac{11}{6}} = \frac{6}{11} x^{\frac{11}{6}} \] Now we evaluate this from \( 0 \) to \( 60 \): \[ C(60) = \left[ \frac{6}{11} x^{\frac{11}{6}} \right]_{0}^{60} = \frac{6}{11} (60^{\frac{11}{6}} - 0^{\frac{11}{6}}) \] Calculating \( 60^{\frac{11}{6}} \): \[ 60^{\frac{11}{6}} = (60^{\frac{1}{6}})^{11} = 60^{1.8333} \approx 1030.4 \quad \text{(using a calculator)} \] Thus, \[ C(60) = \frac{6}{11} \cdot 1030.4 \approx 561.2 \] So, the cost of installing \( 60 \, \text{ft}^2 \) of countertop is approximately **$561.20**. ### c) Find the cost of installing an extra \( 13 \, \text{ft}^2 \) of countertop after \( 60 \, \text{ft}^2 \) have already been installed. To find the cost of installing an additional \( 13 \, \text{ft}^2 \) after \( 60 \, \text{ft}^2 \), we need to evaluate the integral from \( 60 \) to \( 73 \): \[ C(73) - C(60) = \int_{60}^{73} C'(x) \, dx = \int_{60}^{73} x^{\frac{5}{6}} \, dx \] Calculating this integral: \[ \int_{60}^{73} x^{\frac{5}{6}} \, dx = \left[ \frac{6}{11} x^{\frac{11}{6}} \right]_{60}^{73} \] Now we need to calculate \( 73^{\frac{11}{6}} \): \[ 73^{\frac{11}{6}} \approx 1130.4 \quad \text{(using a calculator)} \] Now we can evaluate: \[ C(73) - C(60) = \frac{6}{11} (73^{\frac{11}{6}} - 60^{\frac{11}{6}}) = \frac{6}{11} (1130.4 - 1030.4) = \frac{6}{11} \cdot 100 \] Calculating this gives: \[ C(73) - C(60) = \frac{600}{11} \approx 54.55 \] So, the cost of installing an extra \( 13 \, \text{ft}^2 \) of countertop after \( 60 \, \text{ft}^2 \) have already been installed is approximately **$54.55**.