Questions: Find a model for the Body Mass Index (BMI) of a person, given that BMI varies directly as a person's weight in pounds and inversely as the square of the person's height in inches. If a 6 ft tall person weighing 158 lb has a BMI of 21.43, how much weight would a 6 ft, 4 in. tall person weighing 170 lb need to gain or lose to have a BMI of 20? Write your answer as a positive value for weight gain or a negative value for weight loss. Round your answer to one decimal place.
Transcript text: Find a model for the Body Mass Index (BMI) of a person, given that BMI varies directly as a person's weight in pounds and inversely as the square of the person's height in inches. If a 6 ft tall person weighing 158 lb has a BMI of 21.43 , how much weight would a $6 \mathrm{ft}, 4 \mathrm{in}$. tall person weighing 170 lb need to gain or lose to have a BMI of 20 ? Write your answer as a positive value for weight gain or a negative value for weight loss. Round your answer to one decimal place.
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Solution
Solution Steps
Step 1: Establish the BMI Formula
The Body Mass Index (BMI) is given by the formula:
\[ \text{BMI} = \frac{703 \times \text{weight (lb)}}{\text{height (in)}^2} \]
Step 2: Calculate the Height in Inches
Convert the height of the person from feet and inches to inches.
6 feet = 72 inches
4 inches = 4 inches
Total height = 72 + 4 = 76 inches
Step 3: Set Up the Equation for the Desired BMI
We need to find the weight \( W \) that gives a BMI of 20 for a person who is 76 inches tall.
\[ 20 = \frac{703 \times W}{76^2} \]
Step 4: Solve for the Weight \( W \)
Rearrange the equation to solve for \( W \):
\[ W = \frac{20 \times 76^2}{703} \]
\[ W = \frac{20 \times 5776}{703} \]
\[ W = \frac{115520}{703} \]
\[ W \approx 164.3 \text{ lb} \]
Step 5: Calculate the Weight Change
The current weight of the person is 170 lb. To find the weight change:
\[ \text{Weight change} = 170 \text{ lb} - 164.3 \text{ lb} \]
\[ \text{Weight change} = 5.7 \text{ lb} \]
Final Answer
The person needs to lose approximately 5.7 lb to have a BMI of 20.