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question 1
you want to start an on - campus business making and selling custom - made t - shirts. after some research, you estimate that the total cost of making c(t)=6.5t + 600 dollars. when you set a unit price of p dollars per t - shirt, there will be a demand of t t - shirts, given by the demand equation 75p+2t = 3000.
given this information, what is the maximum amount of revenue that can be made making and selling t - shirts on campus?
round your answer appropriately and input just the number that you get. do not include any labels or units.

Explanation:

Step1: Express $p$ in terms of $T$ from the demand - equation

From $75p + 2T=3000$, we can solve for $p$:
$75p=3000 - 2T$, so $p=\frac{3000 - 2T}{75}=40-\frac{2T}{75}$

Step2: Write the revenue function $R(T)$

Revenue $R(T)=p\times T$. Substitute $p = 40-\frac{2T}{75}$ into the revenue formula:
$R(T)=(40-\frac{2T}{75})T = 40T-\frac{2T^{2}}{75}$

Step3: Find the derivative of the revenue function

The derivative of $R(T)$ with respect to $T$ is $R^\prime(T)=\frac{d}{dT}(40T-\frac{2T^{2}}{75})$.
Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $R^\prime(T)=40-\frac{4T}{75}$

Step4: Find the critical points

Set $R^\prime(T) = 0$ to find the critical points:
$40-\frac{4T}{75}=0$
$\frac{4T}{75}=40$
$T = 750$

Step5: Determine if the critical point is a maximum

Take the second - derivative of $R(T)$: $R^{\prime\prime}(T)=\frac{d}{dT}(40-\frac{4T}{75})=-\frac{4}{75}<0$. Since $R^{\prime\prime}(T)<0$ when $T = 750$, $T = 750$ is a point of maximum.

Step6: Calculate the maximum revenue

Substitute $T = 750$ into the revenue function $R(T)$:
$R(750)=40\times750-\frac{2\times(750)^{2}}{75}$
$R(750)=30000-\frac{2\times562500}{75}$
$R(750)=30000 - 15000$
$R(750)=15000$

Answer:

15000