Math Problem Statement

If the total cost function for producing x lamps is C(x) = 360 + 35x + 0.1x^2 dollars, producing how many units will result in a minimum average cost per unit? Find the minimum average cost. A company handles an apartment building with 50 units. Experience has shown that if the rent for each of the units is $740 per month, all of the units will be filled, but 1 unit will become vacant for each $20 increase in this monthly rate. If the monthly cost of maintaining the apartment building is $12 per rented unit, what rent should be charged per month to maximize the profit?

Solution

Let's break down both questions from the image step by step.

Problem 1: Cost Function Minimization

Given:

  • Total cost function C(x)=360+35x+0.1x2C(x) = 360 + 35x + 0.1x^2

You need to find:

  1. The number of units xx that minimizes the average cost per unit.
  2. The minimum average cost.

Step 1: Average Cost Function

The average cost function AC(x)AC(x) is the total cost C(x)C(x) divided by the number of units xx:

AC(x)=C(x)x=360+35x+0.1x2xAC(x) = \frac{C(x)}{x} = \frac{360 + 35x + 0.1x^2}{x} Simplifying this expression: AC(x)=360x+35+0.1xAC(x) = \frac{360}{x} + 35 + 0.1x

Step 2: Minimizing Average Cost

To find the number of units xx that minimizes the average cost, we need to differentiate the average cost function and set the derivative equal to zero:

AC(x)=360x2+0.1AC'(x) = -\frac{360}{x^2} + 0.1 Set this equal to zero: 360x2+0.1=0-\frac{360}{x^2} + 0.1 = 0 360x2=0.1\frac{360}{x^2} = 0.1 x2=3600.1=3600x^2 = \frac{360}{0.1} = 3600 x=60x = 60

Thus, the number of units that minimizes the average cost is x=60x = 60.

Step 3: Minimum Average Cost

Now, substitute x=60x = 60 back into the average cost function:

AC(60)=36060+35+0.1(60)AC(60) = \frac{360}{60} + 35 + 0.1(60) AC(60)=6+35+6=47dollars/unitAC(60) = 6 + 35 + 6 = 47 \, \text{dollars/unit}

The minimum average cost is $47 per unit.


Problem 2: Maximizing Profit for the Apartment

Given:

  • 50 units in total.
  • Rent of $740 fills all units, but 1 unit becomes vacant for each $20 increase in rent.
  • Monthly cost to maintain the building is $12 per unit.

You need to find the rent that maximizes profit.

Step 1: Revenue Function

Let the number of vacant units be vv. For every $20 increase in rent, 1 unit becomes vacant, so the rent is 740+20v740 + 20v. If there are vv vacant units, the number of rented units is 50v50 - v. The revenue function R(v)R(v) is:

R(v)=(50v)(740+20v)R(v) = (50 - v)(740 + 20v)

Expand this expression: R(v)=(50)(740)+(50)(20v)v(740)v(20v)R(v) = (50)(740) + (50)(20v) - v(740) - v(20v) R(v)=37000+1000v740v20v2R(v) = 37000 + 1000v - 740v - 20v^2 R(v)=37000+260v20v2R(v) = 37000 + 260v - 20v^2

Step 2: Cost Function

The cost to maintain the building is $12 per rented unit. Therefore, the cost function C(v)C(v) is:

C(v)=12(50v)=60012vC(v) = 12(50 - v) = 600 - 12v

Step 3: Profit Function

The profit function P(v)P(v) is revenue minus cost:

P(v)=R(v)C(v)P(v) = R(v) - C(v) P(v)=(37000+260v20v2)(60012v)P(v) = (37000 + 260v - 20v^2) - (600 - 12v) P(v)=36400+272v20v2P(v) = 36400 + 272v - 20v^2

Step 4: Maximizing Profit

To find the maximum profit, differentiate the profit function and set the derivative equal to zero:

P(v)=27240vP'(v) = 272 - 40v Set this equal to zero: 27240v=0272 - 40v = 0 v=27240=6.8v = \frac{272}{40} = 6.8

Since vv must be an integer, round to the nearest integer, v=7v = 7.

Step 5: Calculate Optimal Rent

Now, substitute v=7v = 7 into the rent formula:

Optimal rent=740+20(7)=740+140=880\text{Optimal rent} = 740 + 20(7) = 740 + 140 = 880

The optimal rent to maximize profit is $880.


Would you like more details or have any questions about these solutions?

Here are some related questions:

  1. How do you find the derivative of a function using the power rule?
  2. What are the conditions to check for a maximum or minimum value in a function?
  3. Can you explain how revenue functions and cost functions combine to form profit functions?
  4. How do second-order derivatives help in identifying whether a critical point is a maximum or minimum?
  5. What happens to the average cost function as production increases indefinitely?

Tip: When optimizing costs or profits, always clearly define the objective function (average cost, profit, etc.) before proceeding with differentiation.

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Math Problem Analysis

Mathematical Concepts

Cost minimization
Profit maximization
Differentiation
Revenue function
Quadratic functions

Formulas

C(x) = 360 + 35x + 0.1x^2
AC(x) = C(x) / x
Revenue function: R(v) = (50 - v)(740 + 20v)
Profit function: P(v) = Revenue - Cost

Theorems

Derivative of cost function to find minimum cost
Derivative of profit function to find maximum profit

Suitable Grade Level

Undergraduate or Advanced High School