Questions: 7. The bottom of a box is to be a rectangle with a perimeter of 36 cm. The box must be 4 cm high. What dimensions give the maximum volume? A. I=14 cm, w=4 cm B. I=9 cm, w=9 cm C. I=12 cm, w=6 cm D. I=8 cm, w=10 cm 8. The bottom of a box is to be a rectangle with a perimeter of 36 cm. The box must be 4 cm high. What is the maximum volume? A. 324 cm^3 B. 320 cm^3 C. 288 cm^3 D. 224 cm^3

7. The bottom of a box is to be a rectangle with a perimeter of 36 cm. The box must be 4 cm high. What dimensions give the maximum volume?
A. I=14 cm, w=4 cm
B. I=9 cm, w=9 cm
C. I=12 cm, w=6 cm
D. I=8 cm, w=10 cm
8. The bottom of a box is to be a rectangle with a perimeter of 36 cm. The box must be 4 cm high. What is the maximum volume?
A. 324 cm^3
B. 320 cm^3
C. 288 cm^3
D. 224 cm^3

Transcript text: 7. The bottom of a box is to be a rectangle with a perimeter of 36 cm . The box must be 4 cm high. What dimensions give the maximum volume? A. $\quad I=14 \mathrm{~cm}, w=4 \mathrm{~cm}$ B. $\quad I=9 \mathrm{~cm}, w=9 \mathrm{~cm}$ C. $I=12 \mathrm{~cm}, w=6 \mathrm{~cm}$ D. $\quad I=8 \mathrm{~cm}, w=10 \mathrm{~cm}$ 8. The bottom of a box is to be a rectangle with a perimeter of 36 cm . The box must be 4 cm high. What is the maximum volume? A. $324 \mathrm{~cm}^{3}$ B. $320 \mathrm{~cm}^{3}$ C. $288 \mathrm{~cm}^{3}$ D. $224 \mathrm{~cm}^{3}$

Solution

Solution Steps

Step 1: Define the Problem

We need to find the dimensions of a rectangular box with a perimeter of $ 36 \, \text{cm} $ and a height of $ 4 \, \text{cm} $ that maximize the volume. The volume $ V $ of the box can be expressed as: \[ V = l \cdot w \cdot h \] where $ l $ is the length, $ w $ is the width, and $ h $ is the height.

Step 2: Express Width in Terms of Length

Given the perimeter $ P $ of the rectangle is $ 36 \, \text{cm} $, we can express the width $ w $ in terms of the length $ l $: \[ P = 2l + 2w \implies w = 18 - l \]

Step 3: Write the Volume Function

Substituting $ w $ into the volume formula, we have: \[ V = l \cdot (18 - l) \cdot 4 = 72l - 4l^2 \]

Step 4: Find the Critical Points

To maximize the volume, we take the derivative of $ V $ with respect to $ l $ and set it to zero: \[ \frac{dV}{dl} = 72 - 8l = 0 \implies l = 9 \]

Step 5: Calculate the Maximum Volume

Substituting $ l = 9 $ back into the expression for $ w $: \[ w = 18 - 9 = 9 \] Now, we can calculate the maximum volume: \[ V = 9 \cdot 9 \cdot 4 = 324 \, \text{cm}^3 \]