Question

1. a small independent motion picture company determines the profit p for producing n dvd copies of a recent release is p = -0.02n² + 3.40n - 16. p is the profit in thousands of dollars, and n is the number of units in thousands of units.

a. how many dvds should the company produce to maximize the profit?
b. what will the maximize profit be?

1. a local nursery sells a large number of ornamental trees every year. the owners have determined the cost per tree c for buying and caring for each tree before it is sold is c = 0.001n² - 0.3n + 50. in this function, c is the cost per tree in dollars and n is the number of trees in stock.

a. how many trees will minimize the cost per tree?
b. what will the minimum cost per tree be?

Explanation:

Step1: Identify the function type

The profit function $P = - 0.02n^{2}+3.40n - 16$ is a quadratic function in the form $y = ax^{2}+bx + c$ where $a=-0.02$, $b = 3.40$ and $c=-16$. For a quadratic function $y = ax^{2}+bx + c$, the vertex of the parabola gives the maximum or minimum value. The $x$-coordinate of the vertex is $n=-\frac{b}{2a}$ when $a
eq0$.

Step2: Find the number of DVDs for maximum profit (15a)

Substitute $a=-0.02$ and $b = 3.40$ into the formula $n=-\frac{b}{2a}$.

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Since $n$ is in thousands of units, the company should produce $85000$ DVDs.

Step3: Find the maximum profit (15b)

Substitute $n = 85$ into the profit function $P=-0.02n^{2}+3.40n - 16$.

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So the maximum profit is $128.5$ thousand dollars or $\$128500$.

Step4: For the cost - function of trees (16a)

The cost function $C = 0.001n^{2}-0.3n + 50$ is a quadratic function with $a = 0.001$, $b=-0.3$ and $c = 50$. Using the formula $n=-\frac{b}{2a}$ for the $x$-coordinate of the vertex.

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$$

So $150$ trees will minimize the cost per tree.

Step5: Find the minimum cost per tree (16b)

Substitute $n = 150$ into the cost function $C=0.001n^{2}-0.3n + 50$.

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$$

So the minimum cost per tree is $\$27.5$.

Answer:

15a. $85000$ DVDs
15b. $\$128500$
16a. $150$ trees
16b. $\$27.5$