---
# System prepended metadata

title: In-Class Activity 4.6B
tags: ['4.6', in-class-activity]

---

# Math 132, Activity for Section 4.6



Show your work, explain your solution.

## Question 1:
Find the dimensions of a rectangle with an area of 200 square feet that has a minimum perimeter.


<details> <summary> Example: </summary>

Find the dimensions of a rectangle with an area of 500 square meters that has a minimum perimeter.
    


### Step 1: Define the variables

Let the length of the rectangle be $l$ meters and the width be $w$ meters. The area of the rectangle is given by:

$$
l \cdot w = 500
$$

The perimeter $P$ of the rectangle is:

$$
P = 2l + 2w
$$

We need to minimize the perimeter $P$ subject to the constraint that the area is $500$ square meters.

### Step 2: Express one variable in terms of the other

From the area equation $l \cdot w = 500$, solve for $w$ in terms of $l$:

$$
w = \frac{500}{l}
$$

### Step 3: Substitute into the perimeter equation

Substitute $w = \frac{500}{l}$ into the perimeter equation:

$$
P = 2l + 2 \cdot \frac{500}{l}
$$
$$
P = 2l + \frac{1000}{l}
$$

### Step 4: Minimize the perimeter

To find the value of $l$ that minimizes the perimeter, take the derivative of $P$ with respect to $l$ and set it equal to zero:

$$
\frac{dP}{dl} = 2 - \frac{1000}{l^2}
$$

Set the derivative equal to zero:

$$
2 - \frac{1000}{l^2} = 0
$$

Solve for $l$:

$$
2 = \frac{1000}{l^2}
$$
$$
l^2 = \frac{1000}{2} = 500
$$
$$
l = \sqrt{500}
$$
$$
l = 10\sqrt{5} \ \text{meters}
$$

### Step 5: Find the width

Now, substitute $l = 10\sqrt{5}$ back into the equation for $w$:

$$
w = \frac{500}{l} = \frac{500}{10\sqrt{5}} = 10\sqrt{5} \ \text{meters}
$$

### Step 6: Conclusion

The rectangle that minimizes the perimeter is a square with dimensions:

$$
l = 10\sqrt{5} \ \text{meters}, \quad w = 10\sqrt{5} \ \text{meters}
$$

Thus, the dimensions of the rectangle with an area of 500 square meters that has the minimum perimeter are $10\sqrt{5} \times 10\sqrt{5}$ meters.


</details>


## Question 2:
A 300-room hotel in Las Vegas is filled to capacity every night at $80 per room. For each $1 increase in rent, 3 fewer rooms are rented. If each rented room costs $10 to service per day, how much should the management charge for each room to maximize gross profit? What is the maximum gross profit?

<details> <summary> Example: </summary>



A 400-room hotel is filled to capacity every night at $100 per room. For each $2 increase in rent, 4 fewer rooms are rented. If each rented room costs $15 to service per day, how much should the management charge for each room to maximize gross profit? What is the maximum gross profit?

### Step 1: Define the variables

Let $x$ represent the number of $2$ increases in rent. The price per room as a function of $x$ is:

$$
p(x) = 100 + 2x
$$

The number of rooms rented as a function of $x$ is:

$$
\text{rooms rented}=400 - 4x
$$

### Step 2: Revenue function

The revenue $R(x)$ is the product of the price per room and the number of rooms rented:

$$
R(x) = p(x) \cdot \text{rooms rented} = (100 + 2x)(400 - 4x)
$$

Expand the revenue function:

$$
R(x) = 100(400) + 100(-4x) + 2x(400) + 2x(-4x)
$$
$$
R(x) = 40000 - 400x + 800x - 8x^2
$$
$$
R(x) = 40000 + 400x - 8x^2
$$

### Step 3: Cost function

The total cost $C(x)$ of servicing the rented rooms is based on the cost per room $15 and the number of rooms rented $400-4x$:

$$
C(x) = 15 \cdot \text{rooms rented}=15 (400 - 4x)
$$

Expand the cost function:

$$
C(x) = 15(400) - 15(4x)
$$
$$
C(x) = 6000 - 60x
$$

### Step 4: Profit function

The profit function $P(x)$ is the revenue minus the cost:

$$
P(x) = R(x) - C(x)
$$

Substitute the expressions for $R(x)$ and $C(x)$:

$$
P(x) = (40000 + 400x - 8x^2) - (6000 - 60x)
$$

Simplify:

$$
P(x) = 40000 + 400x - 8x^2 - 6000 + 60x
$$
$$
P(x) = 34000 + 460x - 8x^2
$$

### Step 5: Maximize the profit

To maximize the profit, take the derivative of $P(x)$ with respect to $x$ and set it equal to zero:

$$
\frac{dP}{dx} = 460 - 16x
$$

Set the derivative equal to zero:

$$
460 - 16x = 0
$$

Solve for $x$:

$$
x = \frac{460}{16} = 28.75
$$

### Step 6: Calculate the optimal price and number of rooms

Now that we know $x = 28.75$, calculate the optimal price per room:

$$
p(x) = 100 + 2(28.75) = 100 + 57.5 = 157.5
$$

The number of rooms rented at this price:

$$
400 - 4(28.75) = 400 - 115 = 285 \ \text{rooms}
$$

### Step 7: Calculate the maximum profit

Substitute $x = 28.75$ into the profit function $P(x) = 34000 + 460x - 8x^2$:

$$
P(28.75) = 34000 + 460(28.75) - 8(28.75)^2
$$

Calculate:

$$
P(28.75) = 34000 + 13225 - 6606.25 = 40618.75
$$

### Final Answer:

- The management should charge **$157.50** per room.
- The number of rooms rented will be **285 rooms**.
- The maximum gross profit is **$40,618.75**.



</details>

## Question 3:
Agriculture. A commercial cherry grower estimates from past records that if 30 trees are planted per acre, each tree will yield an average of 50 pounds of cherries per season. If, for each additional tree planted per acre (up to 20), the average yield per tree is reduced by 1 pound, how many trees should be planted per acre to obtain the maximum yield per acre? What is the maximum yield?

<details> <summary> Example: </summary>


Agriculture. A commercial apple grower estimates from past records that if 40 trees are planted per acre, each tree will yield an average of 60 pounds of apples per season. For each additional tree planted per acre (up to 25), the average yield per tree is reduced by 2 pounds. How many trees should be planted per acre to obtain the maximum yield per acre? What is the maximum yield?

### Step 1: Define the variables

Let $x$ represent the number of additional trees planted per acre. Initially, 40 trees are planted, so the total number of trees per acre is:

$$
\text{trees per acre}=40 + x
$$

The average yield per tree starts at 60 pounds, but decreases by 2 pounds for each additional tree. Therefore, the yield per tree as a function of $x$ is:

$$
\text{yield per tree}=60 - 2x
$$

### Step 2: Total yield function

The total yield per acre is the product of the number of trees and the yield per tree. The total yield $Y(x)$ as a function of $x$ is:

$$
Y(x) = (40 + x)(60 - 2x)
$$

### Step 3: Expand the yield function

Expand the expression for $Y(x)$:

$$
Y(x) = 40(60 - 2x) + x(60 - 2x)
$$
$$
Y(x) = 2400 - 80x + 60x - 2x^2
$$
$$
Y(x) = 2400 - 20x - 2x^2
$$

### Step 4: Maximize the yield

To maximize the yield, take the derivative of $Y(x)$ with respect to $x$ and set it equal to zero:

$$
\frac{dY}{dx} = -20 - 4x
$$

Set the derivative equal to zero:

$$
-20 - 4x = 0
$$

Solve for $x$:

$$
4x = -20
$$
$$
x = -5
$$

### Step 5: Interpret the result

The result $x = -5$ implies that the number of trees should be **reduced** by 5 trees from the initial 40 trees. Thus, the optimal number of trees to plant per acre is:

$$
40 - 5 = 35 \ \text{trees}
$$

### Step 6: Calculate the maximum yield

Now, substitute $x = -5$ into the yield function to calculate the maximum yield:

$$
Y(-5) = 2400 - 20(-5) - 2(-5)^2
$$
$$
Y(-5) = 2400 + 100 - 50 = 2450
$$

Thus, the maximum yield is **2450 pounds** of apples per acre.

### Final Answer:

- The grower should plant **35 trees** per acre to maximize yield.
- The maximum yield per acre is **2450 pounds** of apples.



</details>