# A commodity has a demand function modeled by p=126−0.5x and a to

A commodity has a demand function modeled by $$\displaystyle{p}={126}−{0.5}{x}$$ and a total cost function modeled by $$\displaystyle{C}={50}{x}+{39.75}$$, where x is the number of units.
(a)Use the first marginal analysis criterion presented in class to find the production level that yields a maximum profit. Then use the second criterion to ensure that level yields a maximum and not a minimum. Finally, what unit price (in dollars) yields a maximum profit?
$per unit (b)When the profit is maximized, what is the average total cost (in dollars) per unit? (Round your answer to two decimal places.)$ per unit

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Blanche McClain

Step 1
Demand function $$\displaystyle{P}={126}-{0.5}{x}{\left({1}\right)}$$
Total cost function $$\displaystyle{C}={50}{x}+{39.75}$$
a). Revenue R(x) =price x number sold
$$\displaystyle{R}{\left({x}\right)}={P}{x}$$
$$\displaystyle{R}{\left({x}\right)}={\left({126}-{0.5}{x}\right)}{x}$$
$$\displaystyle{R}{\left({x}\right)}={126}{x}-{0.5}{x}^{{{2}}}$$
Step 2
Profit P(x)=revenue-cost
$$\displaystyle={\left({126}{x}-{0.5}{x}^{{{2}}}\right)}-{\left({50}{x}+{39.75}\right)}$$
$$\displaystyle={126}{x}-{0.5}{x}^{{{2}}}-{50}{x}-{39.75}$$
$$\displaystyle=-{0.5}{x}^{{{2}}}+{76}{x}-{39.75}$$
To maximize profit P'(x)=0
$$\displaystyle{P}'{\left({x}\right)}=-{0.5}{\left({2}{x}\right)}+{76}$$
$$\displaystyle{P}'{\left({x}\right)}=-{x}+{76}={0}$$
x=76
and $$\displaystyle{P}{''''}{\left({x}\right)}=-{1}{<}{0}$$ (maximum)
first and second derivative shows that at x=76, P(x) is maximum.

###### Not exactly what you’re looking for?
Huses1969
Step 1
b
Average cost $$\displaystyle\overline{{{C}}}{\left({x}\right)}={\frac{{{C}{\left({x}\right)}}}{{{x}}}}={\frac{{{50}{x}+{39.75}}}{{{x}}}}$$
substitute the value of x=76, we get
$$\displaystyle\overline{{{C}}}{\left({x}\right)}=\overline{{{C}}}{\left({76}\right)}={\frac{{{50}{\left({76}\right)}+{39.75}}}{{{76}}}}$$
$$\displaystyle\overline{{{C}}}{\left({x}\right)}=\{50.52}$$ per unit