Answer :
To solve for the dimensions of the stage with an area of 280 square feet, we need to use the given expressions for its length and width.
1. Set up the equation:
The area of the stage is calculated by multiplying the length and the width. We are given that:
- Length \( = x - 12 \)
- Width \( = x - 18 \)
The area of the stage is then:
[tex]\[ (x - 12)(x - 18) = 280 \][/tex]
2. Expand and simplify the equation:
Expanding the left-hand side:
[tex]\[ x^2 - 18x - 12x + 216 = 280 \][/tex]
Simplifying further:
[tex]\[ x^2 - 30x + 216 = 280 \][/tex]
3. Move all terms to one side to form a quadratic equation:
Subtract 280 from both sides:
[tex]\[ x^2 - 30x + 216 - 280 = 0 \][/tex]
Simplify:
[tex]\[ x^2 - 30x - 64 = 0 \][/tex]
4. Solve the quadratic equation:
To find the value(s) of \(x\), solve the quadratic equation \(x^2 - 30x - 64 = 0\). The solutions to this equation are:
[tex]\[ x = -2 \quad \text{or} \quad x = 32 \][/tex]
5. Find the corresponding dimensions:
We use the values of \(x\) to find the corresponding dimensions.
- For \( x = -2 \):
- Length \( = -2 - 12 = -14 \) (not a valid physical dimension)
- Width \( = -2 - 18 = -20 \) (not a valid physical dimension)
Since both dimensions are negative, this value of \(x\) is not physically meaningful for our problem.
- For \( x = 32 \ ):
- Length \( = 32 - 12 = 20 \)
- Width \( = 32 - 18 = 14 \)
These dimensions are both positive and thus valid.
6. Conclusion:
Therefore, the valid dimensions for the stage are:
- Length = 20 feet
- Width = 14 feet
1. Set up the equation:
The area of the stage is calculated by multiplying the length and the width. We are given that:
- Length \( = x - 12 \)
- Width \( = x - 18 \)
The area of the stage is then:
[tex]\[ (x - 12)(x - 18) = 280 \][/tex]
2. Expand and simplify the equation:
Expanding the left-hand side:
[tex]\[ x^2 - 18x - 12x + 216 = 280 \][/tex]
Simplifying further:
[tex]\[ x^2 - 30x + 216 = 280 \][/tex]
3. Move all terms to one side to form a quadratic equation:
Subtract 280 from both sides:
[tex]\[ x^2 - 30x + 216 - 280 = 0 \][/tex]
Simplify:
[tex]\[ x^2 - 30x - 64 = 0 \][/tex]
4. Solve the quadratic equation:
To find the value(s) of \(x\), solve the quadratic equation \(x^2 - 30x - 64 = 0\). The solutions to this equation are:
[tex]\[ x = -2 \quad \text{or} \quad x = 32 \][/tex]
5. Find the corresponding dimensions:
We use the values of \(x\) to find the corresponding dimensions.
- For \( x = -2 \):
- Length \( = -2 - 12 = -14 \) (not a valid physical dimension)
- Width \( = -2 - 18 = -20 \) (not a valid physical dimension)
Since both dimensions are negative, this value of \(x\) is not physically meaningful for our problem.
- For \( x = 32 \ ):
- Length \( = 32 - 12 = 20 \)
- Width \( = 32 - 18 = 14 \)
These dimensions are both positive and thus valid.
6. Conclusion:
Therefore, the valid dimensions for the stage are:
- Length = 20 feet
- Width = 14 feet
