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) \(\displaystyle{R}=-{0.01}{N}^{2}+{55}{N}\)

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) \(\displaystyle{R}=-{0.01}{N}^{2}+{55}{N}\)