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The marginal cost of a product is modeled by \frac{dC}{dx}= \frac{16}{\sqrt[3] {16x + 3}} where x...

Question:

The marginal cost of a product is modeled by {eq}\frac{dC}{dx}= \frac{16}{\sqrt[3] {16x + 3}} {/eq} where x is the number of units. When x = 17, C = 120.

(a) Find the cost function. (Round your constant term to two decimal places.)

c =

(b) Find the cost of producing 90 units. (Round your answer to two decimal places.)

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Integration

There are many ways to integrate a function depending on the condition. Integration is the process of calculating the sum of the infinitesimal elements of a function in a certain limit. Integration is also performed to solve the differential equations.

Answer and Explanation:


Given data

  • The marginal cost of a function is: {eq}\dfrac{{dC}}{{dx}} = \dfrac{{16}}{{\sqrt[3]{{16x + 3}}}} {/eq}
  • The number of units is: {eq}x = 17 {/eq}
  • The cost according to above number of units is: {eq}C = 120 {/eq}


(a)

Integrate the above marginal cost function.

{eq}\begin{align*} \int {dC} &= \int {\dfrac{{16}}{{\sqrt[3]{{16x + 3}}}}dx} \\ C &= 16\int {{{\left( {16x + 3} \right)}^{\dfrac{{ - 1}}{3}}}dx} \\ C &= 16\left( {\dfrac{{{{\left( {16x + 3} \right)}^{\dfrac{2}{3}}}}}{{\dfrac{2}{3}}}\left( {\dfrac{1}{{16}}} \right)} \right) + K\\ C &= \dfrac{3}{2}\left( {{{\left( {16x + 3} \right)}^{\dfrac{2}{3}}}} \right) + K\cdots\cdots\rm{(I)} \end{align*} {/eq}


Apply the given conditions.

{eq}\begin{align*} 120 &= \dfrac{3}{2}\left( {{{\left( {16\left( {17} \right) + 3} \right)}^{\dfrac{2}{3}}}} \right) + K\\ 120 &= 63.432 + K\\ K &= 56.568 \end{align*} {/eq}


Substitute the known value in equation (I).

{eq}C = \dfrac{3}{2}\left( {{{\left( {16x + 3} \right)}^{\dfrac{2}{3}}}} \right) + 56.568 {/eq}


Thus, the cost function is {eq}C = \dfrac{3}{2}\left( {{{\left( {16x + 3} \right)}^{\dfrac{2}{3}}}} \right) + 56.568 {/eq}.


(b)

  • The number of units is: {eq}x = 90 {/eq}


Substitute the known value in cost function.

{eq}\begin{align*} C &= \dfrac{3}{2}\left( {{{\left( {16\left( {90} \right) + 3} \right)}^{\dfrac{2}{3}}}} \right) + 56.568\\ &= 191.544 + 56.568\\ &= 248.112 \end{align*} {/eq}


Thus, the cost is {eq}248.112 {/eq}.


Learn more about this topic:

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Integration Problems in Calculus: Solutions & Examples

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