Jason builds doghouses for a pet store. Each doghouse is a wooden structure with a rectangular base that has an area of 21 square feet and a length that is 4 feet more than its width.
The Correct Answer and Explanation is :
To solve the problem, we need to determine the dimensions of the rectangular base of the doghouse that Jason is building. We know two key pieces of information:
- The area of the rectangular base is 21 square feet.
- The length of the rectangle is 4 feet more than its width.
Let’s denote the width of the doghouse as ( w ) (in feet). According to the problem, the length ( l ) can be expressed as:
[
l = w + 4
]
Since the area ( A ) of a rectangle is given by the formula:
[
A = l \times w
]
Substituting the expression for length into the area formula, we have:
[
21 = (w + 4) \times w
]
Expanding this equation gives us:
[
21 = w^2 + 4w
]
Rearranging the equation leads us to a standard quadratic form:
[
w^2 + 4w – 21 = 0
]
Now we can solve this quadratic equation using the quadratic formula:
[
w = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
]
Here, ( a = 1 ), ( b = 4 ), and ( c = -21 ). Plugging in these values:
[
w = \frac{-4 \pm \sqrt{4^2 – 4 \cdot 1 \cdot (-21)}}{2 \cdot 1}
]
Calculating the discriminant:
[
4^2 – 4 \cdot 1 \cdot (-21) = 16 + 84 = 100
]
Now substituting back into the formula:
[
w = \frac{-4 \pm 10}{2}
]
Calculating the two potential values for ( w ):
- ( w = \frac{6}{2} = 3 )
- ( w = \frac{-14}{2} = -7 ) (not a valid solution since width cannot be negative)
Thus, ( w = 3 ) feet. To find the length:
[
l = w + 4 = 3 + 4 = 7 \text{ feet}
]
In conclusion, the dimensions of the doghouse are 3 feet in width and 7 feet in length. This solution ensures that the area remains 21 square feet, as required. This exercise demonstrates the application of algebraic techniques to solve practical problems involving geometric dimensions, which is essential in fields like construction and design.