A housing committee wants a rectangular swimming pool and surrounding deck with a total area of . The
deck is 3 m wide at each end of the pool and 2 m wide on each side. Find the dimensions of the pool that has the
maximum area.

Question

A housing committee wants a rectangular swimming pool and surrounding deck with a total area of

. The
deck is 3 m wide at each end of the pool and 2 m wide on each side. Find the dimensions of the pool that has the
maximum area.

UpStudy AI · Accepted Answer

We start by letting the length and width of the pool be $$ L $$ and $$ W $$, respectively. The deck surrounds the pool with a width of 3 m at each end (adding 6 m to the length) and 2 m on each side (adding 4 m to the width). This means the overall dimensions (pool plus deck) are:
$$
L + 6 \quad \text{and} \quad W + 4.
$$

The total area (pool plus deck) is given as 150 m$$^2$$, so we have the constraint:
$$
(L+6)(W+4) = 150.
$$

Our goal is to maximize the pool area
$$
A = L \cdot W,
$$
subject to the area constraint above.

### Step 1. Express $$W$$ in Terms of $$L$$

From the constraint:
$$
(L+6)(W+4) = 150 \quad \Longrightarrow \quad W + 4 = \frac{150}{L+6}.
$$
Subtract 4 from both sides to get:
$$
W = \frac{150}{L+6} - 4.
$$

### Step 2. Write the Area of the Pool as a Function of $$L$$

The pool area $$ A $$ becomes:
$$
A(L) = L\left(\frac{150}{L+6} - 4\right).
$$
Simplify:
$$
A(L) = \frac{150L}{L+6} - 4L.
$$

### Step 3. Find the Critical Points by Differentiating $$A(L)$$

Differentiate $$ A(L) $$ with respect to $$ L $$. First, write the derivative of the first term using the quotient rule:
$$
\frac{d}{dL}\left(\frac{150L}{L+6}\right) = \frac{150(L+6) - 150L}{(L+6)^2} = \frac{150 \cdot 6}{(L+6)^2} = \frac{900}{(L+6)^2}.
$$
The derivative of the second term $$-4L$$ is $$-4$$.

Thus, the derivative of $$ A(L) $$ is:
$$
A'(L) = \frac{900}{(L+6)^2} - 4.
$$

Set the derivative equal to zero for maximization:
$$
\frac{900}{(L+6)^2} - 4 = 0.
$$

Solve for $$ L $$:
$$
\frac{900}{(L+6)^2} = 4 \quad \Longrightarrow \quad (L+6)^2 = \frac{900}{4} = 225.
$$

Taking the positive square root (since lengths must be positive):
$$
L + 6 = 15 \quad \Longrightarrow \quad L = 15 - 6 = 9.
$$

### Step 4. Compute $$W$$ Using the Constraint

Substitute $$ L = 9 $$ into the expression for $$ W $$:
$$
W = \frac{150}{9+6} - 4 = \frac{150}{15} - 4 = 10 - 4 = 6.
$$

### Final Answer

The dimensions of the pool that give the maximum area are:
$$
\text{Length } L = 9 \text{ m}, \quad \text{Width } W = 6 \text{ m}.
$$
The pool should be 9 meters in length and 6 meters in width to maximize the area.

A housing committee wants a rectangular swimming pool and surrounding deck with a total area of . The
deck is 3 m wide at each end of the pool and 2 m wide on each side. Find the dimensions of the pool that has the
maximum area.

Upstudy AI Solution

Answer

Solution

Step 1. Express in Terms of

Step 2. Write the Area of the Pool as a Function of

Step 3. Find the Critical Points by Differentiating

Step 4. Compute Using the Constraint

Final Answer

Bonus Knowledge

Related Questions

Latest Calculus Questions

A housing committee wants a rectangular swimming pool and surrounding deck with a total area of . The deck is 3 m wide at each end of the pool and 2 m wide on each side. Find the dimensions of the pool that has the maximum area.

Upstudy AI Solution

Answer

Solution

Step 1. Express in Terms of

Step 2. Write the Area of the Pool as a Function of

Step 3. Find the Critical Points by Differentiating

Step 4. Compute Using the Constraint

Final Answer

Bonus Knowledge

Related Questions

Latest Calculus Questions

A housing committee wants a rectangular swimming pool and surrounding deck with a total area of . The
deck is 3 m wide at each end of the pool and 2 m wide on each side. Find the dimensions of the pool that has the
maximum area.