Question5.4 Systems of Nonlinear Equations in Two Variables
Question 16 of 16 (2 points) I Question Attempt: 1 of Unlimited
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Find the dimensions of a rectangle whose perimeter is 48 m and whose area is . Give the answer in simplest form.
The rectangle is m by m.
Studdy Solution
STEP 1
1. The rectangle has a perimeter of meters.
2. The area of the rectangle is square meters.
3. Let the length of the rectangle be meters and the width be meters.
STEP 2
1. Set up the equations for perimeter and area.
2. Solve the system of equations.
3. Verify the solution.
STEP 3
Set up the equations for perimeter and area:
1. Perimeter equation:
$ 2l + 2w = 48
\]
2. Area equation:
$ lw = 140
\]
STEP 4
Solve the system of equations:
1. Simplify the perimeter equation:
$ l + w = 24
\]
2. Express in terms of :
$ l = 24 - w
\]
3. Substitute in the area equation:
$ (24 - w)w = 140
\]
4. Expand and rearrange the equation:
$ 24w - w^2 = 140
\]
5. Rearrange to form a quadratic equation:
$ w^2 - 24w + 140 = 0
\]
6. Solve the quadratic equation using the quadratic formula:
$ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -24 \), \( c = 140 \).
7. Calculate the discriminant:
$ b^2 - 4ac = 576 - 560 = 16
\]
8. Calculate the roots:
w = \frac{24 \pm \sqrt{16}}{2}
\]
w = \frac{24 \pm 4}{2}
\]
9. The possible values for are:
$ w = \frac{28}{2} = 14 \quad \text{or} \quad w = \frac{20}{2} = 10
\]
10. Corresponding values:
- If , then .
- If , then .
STEP 5
Verify the solution:
1. Check the perimeter:
$ 2(14) + 2(10) = 28 + 20 = 48
\]
2. Check the area:
$ 14 \times 10 = 140
\]
Both conditions are satisfied.
The dimensions of the rectangle are m by m.
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