Question

The cost, C, to produce b baseball bats per day is modeled by the function C(b) = 0.06b2 – 7.2b + 390. What number of bats should be produced to keep costs at a minimum?

Answer 1

To find the number of bats that should be produced to minimize costs, we need to find the minimum point on the cost curve given by the function C(b) = 0.06b^2 - 7.2b + 390. Using the vertex formula, we find that the minimum occurs at b = 60.

Step-by-step explanation:

To find the number of bats that should be produced to keep costs at a minimum, we need to determine the minimum point on the cost curve given by the function C(b) = 0.06b^2 - 7.2b + 390. The minimum point of a quadratic function can be found using the vertex formula: b = -b / (2a), where a is the coefficient of the quadratic term and b is the coefficient of the linear term. In this case, a = 0.06 and b = -7.2. Plugging these values into the formula, we get b = -(-7.2) / (2 * 0.06) = 60.

Therefore, the number of bats that should be produced to keep costs at a minimum is 60.

Answer 2

Check the picture below, that's just an example of a parabola opening upwards.

so the cost equation C(b), which is a quadratic with a positive leading term's coefficient, has the graph of a parabola like the one in the picture, so the cost goes down and down and down, reaches the vertex or namely the minimum, and then goes back up.

bearing in mind that the quantity will be on the x-axis and the cost amount is over the y-axis, what are the coordinates of the vertex of this parabola? namely, at what cost for how many bats?

`\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ C(b) = \stackrel{\stackrel{a}{\downarrow }}{0.06}b^2\stackrel{\stackrel{b}{\downarrow }}{-7.2}b\stackrel{\stackrel{c}{\downarrow }}{+390} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

`\bf \left( -\cfrac{-7.2}{2(0.06)}~~,~~390-\cfrac{(-7.2)^2}{4(0.06)} \right)\implies (60~~,~~390-216) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\textit{number of bats}}{60}~~,~~\stackrel{\textit{total cost}}{174})~\hfill`