The graph to the right shows a region of feasible solutions. Use this region to find maximum and minimum values of the given objective functions, and the locations of these values on the graph. (a) \( z=4 x+3 y \) (b) \( z=2.25 x+0.50 y \)

La derivada de la función \[ \begin{array}{l}f(x)=2 x^{4}+x^{3}-x^{2}+4 \\ \text { a. } f^{\prime}(x)=20 x^{3}+15 x^{2}+34 x-3 \\ f^{\prime}(x)=8 x^{3}+3 x^{2}-2 x \\ f^{\prime}(x)=-2\end{array} \] \( \begin{array}{l}\text { b. }\end{array} \)

3. \( f(x)=\frac{x}{x+3} \)

La derivada de la función \( f(x)=-2 x \) a. \( f^{\prime}(x)=-4 x \) b. \( f^{\prime}(x)=20 x^{3}+15 x^{2}+34 x-3 \) oc. \( f^{\prime}(x)=8 x^{3}+3 x^{2}-2 x \)

La ecuación de la parábola con vértice en el origen y foco en el punto \( (0,-5) \), es: a. \( x^{2}=-20 y \) \( b . x^{2}=-5 y \) \( c y^{2}=20 x \) \( v^{2}=5 v \)

WORKSHEET: ADolications of Integration For a manufacturer, the marginal cost (in dollars) per unit, \( C^{\prime}(x) \), when producing units is given by \( C^{\prime}(x)=3 x^{2}-60 x+400 \). (a) Given that the total cost of producing the first 2 units is \( \$ 900 \), use integration (b) Herme calculate the total cost of producing the first 5 units.