Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 409: 47

$$\eqalign{ & \left( {\text{a}} \right)\$ 69859.33 \cr & \left( {\text{b}} \right)\$ 139583.33 \cr} $$

$$\eqalign{ & {\text{The marginal cost is given by }} \cr & C'\left( x \right) = 300 + 10x - 0.01{x^2} \cr & \cr & {\text{From the definition of the marginal cost }}\left( {{\text{page 177}}} \right){\text{ we have}} \cr & {\text{that the cost function }}C\left( x \right){\text{ is the cost required to produce }}x \cr & {\text{units of a product}}{\text{.}} \cr & \cr & {\text{Using the Theorem 6}}{\text{.3}}{\text{, }}\left( {{\text{Net change and future value}}} \right) \cr & Q\left( b \right) - Q\left( a \right) = \int_a^b {Q'\left( t \right)} dt \cr & \cr & {\text{Then applying the concept theorem}}{\text{, the cost in dollars of}} \cr & {\text{producion is given by}} \cr & C\left( b \right) - C\left( a \right) = \int_a^b {C'\left( t \right)} dt \cr & \cr & \left( {\bf{a}} \right){\text{ }}a = 100{\text{ units}}{\text{, b}} = 150{\text{ units}} \cr & C\left( {150} \right) - C\left( {100} \right) = \int_{100}^{150} {\left( {300 + 10x - 0.01{x^2}} \right)} dx \cr & {\text{Integrating}} \cr & = \left[ {300x + 5{x^2} - \frac{{0.01}}{3}{x^3}} \right]_{100}^{150} \cr & = \left[ {300\left( {150} \right) + 5{{\left( {150} \right)}^2} - \frac{{0.01}}{3}{{\left( {150} \right)}^3}} \right] \cr & - \left[ {300\left( {100} \right) + 5{{\left( {100} \right)}^2} - \frac{{0.01}}{3}{{\left( {100} \right)}^3}} \right] \cr & {\text{Simplifying}} \cr & = 146250 - 76666.66 \cr & = \$ 69859.33 \cr & \cr & \left( {\bf{b}} \right){\text{ }}a = 500{\text{ units}}{\text{, b}} = 550{\text{ units}} \cr & C\left( {550} \right) - C\left( {500} \right) = \int_{500}^{550} {\left( {300 + 10x - 0.01{x^2}} \right)} dx \cr & {\text{Integrating}} \cr & = \left[ {300x + 5{x^2} - \frac{{0.01}}{3}{x^3}} \right]_{500}^{550} \cr & = \left[ {300\left( {550} \right) + 5{{\left( {550} \right)}^2} - \frac{{0.01}}{3}{{\left( {550} \right)}^3}} \right] \cr & - \left[ {300\left( {500} \right) + 5{{\left( {500} \right)}^2} - \frac{{0.01}}{3}{{\left( {500} \right)}^3}} \right] \cr & {\text{Simplifying}} \cr & = 1122916.667 - 983333.33 \cr & = \$ 139583.33 \cr} $$