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# In general, the highest price p per unit of an item at which a manufacturer can sell N items is not constant but is, rather, a function of N. Suppose the manufacturer of widgets has developed the following table showing the highest price p, in dollars, of a widget at which N widgets can be sold. begin{array}{|c|c|} hline Number N & Price p hline 250 & 52.50 hline300 & 52.00hline 350 & 51.50 hline 400 & 51.00 hline end{array} (a) Find a formula for p in terms of N modeling the data in the table. displaystyle{p}= (b) Use a formula to express the total monthly revenue R, in dollars, of this manufacturer in a month as a function of the number N of widgets produced in a month. displaystyle{R}=

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Modeling data distributions
asked 2021-02-25
In general, the highest price p per unit of an item at which a manufacturer can sell N items is not constant but is, rather, a function of N. Suppose the manufacturer of widgets has developed the following table showing the highest price p, in dollars, of a widget at which N widgets can be sold.
$$\begin{array}{|c|c|} \hline Number\ N & Price\ p\\ \hline 250 & 52.50\\ \hline300 & 52.00\\\hline 350 & 51.50\\ \hline 400 & 51.00\\ \hline \end{array}$$
(a) Find a formula for p in terms of N modeling the data in the table.
$$\displaystyle{p}=$$
(b) Use a formula to express the total monthly revenue R, in dollars, of this manufacturer in a month as a function of the number N of widgets produced in a month.
$$\displaystyle{R}=$$

## Answers (1)

2021-02-26
Slope of line joining the points $$\displaystyle{\left({250},{52.50}\right)}{\quad\text{and}\quad}{\left({300},{52.00}\right)}{i}{s}{m}=\frac{{{y}{2}-{y}{1}}}{{{x}{2}-{x}{1}}}=\frac{{{52.00}-{52.50}}}{{{300}-{250}}}=-{0.01}$$
Slope of line joining the points $$\displaystyle{\left({300},{52.00}\right)}{\quad\text{and}\quad}{\left({350},{1.505}\right)}{i}{s}{m}=\frac{{{y}{2}-{y}{1}}}{{{x}{2}-{x}{1}}}=\frac{{{51.50}-{52.00}}}{{{350}-{300}}}=-{0.01}$$ and so on.
Since slope of line joining the points (250, 52.50) and (300, 52.00) is equal to the slope of line joining the points (300, 52.00) and (350, 1.505) , so there is a linear relation between p and N
Let $$\displaystyle{p}={m}{N}+{b}$$
where $$\displaystyle{m}=-{0.01}$$
Therefore,
$$\displaystyle{p}=-{0.01}{N}+{b}$$
Now, the point (250, 52.50) lie on the line, so
$$\displaystyle{52.50}=-{0.01}\times{250}+{b}$$
$$\displaystyle{52.50}=-{2.5}+{b}$$
$$\displaystyle{b}={55}$$
Thus, $$\displaystyle{p}=-{0.01}{N}+{55}$$
(b) The total monthly revenue R, in dollars, of this manufacturer in a month as a function of the number N of widgets produced in a month is determined as follows
$$\displaystyle{R}={p}{N}$$
$$\displaystyle{R}={\left(-{0.01}{N}+{55}\right)}{N}$$
$$\displaystyle{R}=-{0.01}{N}^{2}+{55}{N}$$
Finally answer:
$$\displaystyle{\left({a}\right)}{p}=-{0.01}{N}+{55}$$
(b) $$\displaystyle{R}=-{0.01}{N}^{2}+{55}{N}$$

### Relevant Questions

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Suppose the manufacturer of widgets has developed the following table showing the highest price p, in dollars, of a widget at which N widgets can be sold.
$$\begin{array}{|c|c|} \hline Number\ N & Price\ p\\ \hline 200 & 53.00\\ \hline 250 & 52.50\\\hline 300 & 52.00\\ \hline 350 & 51.50\\ \hline \end{array}$$
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$$\displaystyle{p}=$$
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$$A. 20602060xf(x)$$
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$$B. 20602060xf(x)$$
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$$\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$
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asked 2021-05-05

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