A company manufactures and sells $$ x $$ television sets per month. The monthly cost and price-demand equations are $$ C(x)=75,000+70 x $$ and $$ p(x)=200-\frac{x}{30}, 0 \leq x \leq 6000 $$. (A) Find the maximum revenue. (B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set. (A) The maximum revenue is $$ \$ \square $$ (Type an integer or a decimal.) (B) The maximum profit is $$ \$ \square $$ when $$ \square $$ sets are manufactured and sold for $$ \$ \square $$ each (Type integers or decimals.)

Question

UpStudy AI · Accepted Answer

We start by writing the formulas given in the problem:

- Cost function: 
  $$
  C(x) = 75000 + 70x
  $$
- Price-demand function:
  $$
  p(x)=200-\frac{x}{30}
  $$
- Revenue function (price times number of sets): 
  $$
  R(x)=x\left(200-\frac{x}{30}\right)=200x-\frac{x^2}{30}
  $$

### (A) Finding the Maximum Revenue

1. **Revenue function:**
   $$
   R(x)=200x-\frac{x^2}{30}
   $$

2. **Differentiate $$ R(x) $$ with respect to $$ x $$:**
   $$
   R'(x)=200-\frac{2x}{30}=200-\frac{x}{15}
   $$

3. **Set the derivative equal to zero to find the critical point:**
   $$
   200-\frac{x}{15}=0
   $$
   Solving for $$ x $$:
   $$
   \frac{x}{15}=200 \quad \Longrightarrow \quad x=200\cdot 15=3000
   $$

4. **Calculate the maximum revenue:**
   $$
   R(3000)=200(3000)-\frac{3000^2}{30}
   $$
   First, compute each term:
   $$
   200(3000)=600000
   $$
   $$
   3000^2=9000000 \quad \Longrightarrow \quad \frac{9000000}{30}=300000
   $$
   Hence:
   $$
   R(3000)=600000-300000=300000
   $$

The maximum revenue is $$\boxed{300000}$$.

### (B) Finding the Maximum Profit

1. **Profit function definition:**
   $$
   P(x)= R(x)-C(x)
   $$
   Substitute for $$ R(x) $$ and $$ C(x) $$:
   $$
   P(x)=\left(200x-\frac{x^2}{30}\right)-(75000+70x)
   $$
   Simplify:
   $$
   P(x)=200x-\frac{x^2}{30}-75000-70x
   $$
   $$
   P(x)=130x-\frac{x^2}{30}-75000
   $$

2. **Differentiate $$ P(x) $$ with respect to $$ x $$:**
   $$
   P'(x)=130-\frac{2x}{30}=130-\frac{x}{15}
   $$

3. **Set the derivative equal to zero to find the production level for maximum profit:**
   $$
   130-\frac{x}{15}=0
   $$
   Solve for $$ x $$:
   $$
   \frac{x}{15}=130 \quad \Longrightarrow \quad x=130\cdot 15=1950
   $$

4. **Calculate the maximum profit $$ P(1950) $$:**
   $$
   P(1950)=130(1950)-\frac{1950^2}{30}-75000
   $$
   First, compute each term:
   $$
   130(1950)=253500
   $$
   Next, compute $$ 1950^2 $$:
   $$
   1950^2=3802500 \quad \Longrightarrow \quad \frac{3802500}{30}=126750
   $$
   Now substitute back:
   $$
   P(1950)=253500-126750-75000
   $$
   $$
   P(1950)=253500-201750=51750
   $$

5. **Determine the price to charge:**
   Use the price-demand function:
   $$
   p(1950)=200-\frac{1950}{30}
   $$
   Compute:
   $$
   \frac{1950}{30}=65
   $$
   $$
   p(1950)=200-65=135
   $$

Thus, the maximum profit is $$\boxed{51750}$$ dollars when $$\boxed{1950}$$ sets are manufactured and sold. The company should charge $$\boxed{135}$$ dollars per television set.

Final answers:

- (A) The maximum revenue is $$\displaystyle 300000$$.
- (B) The maximum profit is $$\displaystyle 51750$$ when $$\displaystyle 1950$$ sets are manufactured and sold for $$\displaystyle 135$$ each.
- (A) The maximum revenue is \$300,000. - (B) The maximum profit is \$51,750 when 1,950 sets are manufactured and sold for \$135 each.

Upstudy AI Solution

Answer

Solution

(A) Finding the Maximum Revenue

(B) Finding the Maximum Profit

Beyond the Answer

Part (A): Maximum Revenue

Part (B): Maximum Profit

Final Answers:

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