suppose that the dollar cost of producing x appliances is c(x)=1000 + 70x - 0.1x^2. a. find the average cost…

suppose that the dollar cost of producing x appliances is c(x)=1000 + 70x - 0.1x^2. a. find the average cost per appliance of producing the first 90 appliances. b. find the marginal cost when 90 appliances are produced. c. show that the marginal cost when 90 appliances are produced is approximately the cost of producing one more. the average cost per appliance of producing the first 90 appliances is $ /appliance. (round to the nearest cent as needed.)
Answer
Explanation:
Step1: Calculate total cost for 90 appliances
Substitute $x = 90$ into $c(x)=1000 + 70x-0.1x^{2}$. $c(90)=1000 + 70\times90-0.1\times90^{2}$ $=1000 + 6300-0.1\times8100$ $=1000 + 6300 - 810$ $=6490$
Step2: Calculate average cost per appliance
The average cost $\bar{c}(x)=\frac{c(x)}{x}$. For $x = 90$, $\bar{c}(90)=\frac{c(90)}{90}$. $\bar{c}(90)=\frac{6490}{90}\approx72.11$
Step3: Find the derivative of the cost - function
The derivative of $c(x)=1000 + 70x-0.1x^{2}$ using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ and $\frac{d}{dx}(constant)=0$. $c^\prime(x)=\frac{d}{dx}(1000)+\frac{d}{dx}(70x)-\frac{d}{dx}(0.1x^{2})$ $c^\prime(x)=0 + 70-0.2x$
Step4: Calculate the marginal cost at $x = 90$
Substitute $x = 90$ into $c^\prime(x)$. $c^\prime(90)=70-0.2\times90$ $=70 - 18$ $=52$
Step5: Interpret the marginal cost
The marginal cost $c^\prime(x)$ represents the approximate cost of producing one additional unit when $x$ units are already produced. When $x = 90$, $c^\prime(90) = 52$, which means the cost of producing the 91 - st appliance is approximately $$52$.
Answer:
a. $$72.11$ b. $$52$ c. The marginal cost $c^\prime(90) = 52$ represents the approximate cost of producing one more (the 91 - st) appliance when 90 appliances have been produced.