Question

a persons rectangular dog pen for his dog must have an area of 400 square feet. also, the length must be 20 feet longer than the width. find the dimensions of the pen.
the width of the pen is and the length of the pen is
(simplify your answers, including any radicals. use integers or frac bers in the expressions.)
ft.
cu ft.
sq ft.

Explanation:

Step1: Define Variables

Let the width of the pen be \( w \) feet. Then the length \( l = w + 20 \) feet (since length is 20 feet longer than width).

Step2: Use Area Formula

The area of a rectangle is \( A = l \times w \). Given \( A = 400 \) sq ft, substitute \( l \) and \( A \):
\[ w(w + 20) = 400 \]
Expand the left side:
\[ w^2 + 20w - 400 = 0 \]

Step3: Solve Quadratic Equation

Use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 20 \), \( c = -400 \):
\[ w = \frac{-20 \pm \sqrt{20^2 - 4(1)(-400)}}{2(1)} = \frac{-20 \pm \sqrt{400 + 1600}}{2} = \frac{-20 \pm \sqrt{2000}}{2} \]
Simplify \( \sqrt{2000} = \sqrt{100 \times 20} = 10\sqrt{20} = 10 \times 2\sqrt{5} = 20\sqrt{5} \):
\[ w = \frac{-20 \pm 20\sqrt{5}}{2} = -10 \pm 10\sqrt{5} \]
Since width can't be negative, take the positive root:
\[ w = -10 + 10\sqrt{5} = 10(\sqrt{5} - 1) \approx 10(2.236 - 1) = 12.36 \text{ (but keep exact form for now)} \]
Wait, wait—wait, maybe I made a mistake. Wait, let's check the quadratic again. Wait, \( w^2 + 20w - 400 = 0 \). Wait, discriminant is \( 20^2 + 1600 = 400 + 1600 = 2000 \), correct. But maybe there's a miscalculation. Wait, no—wait, actually, let's solve it properly.

Wait, \( w = \frac{-20 + \sqrt{400 + 1600}}{2} = \frac{-20 + \sqrt{2000}}{2} = \frac{-20 + 10\sqrt{20}}{2} = \frac{-20 + 20\sqrt{5}}{2} = -10 + 10\sqrt{5} \). Yes, that's correct. Then length is \( l = w + 20 = -10 + 10\sqrt{5} + 20 = 10 + 10\sqrt{5} = 10(\sqrt{5} + 1) \).

Wait, but let's check if this is correct. Let's compute \( w \times l \): \( (-10 + 10\sqrt{5})(10 + 10\sqrt{5}) = 10\sqrt{5} \times 10\sqrt{5} - 10 \times 10 + 10\sqrt{5} \times 10 - 10 \times 10\sqrt{5} \)? Wait, no, use difference of squares: \( (a - b)(a + b) = a^2 - b^2 \), but here it's \( (10\sqrt{5} - 10)(10\sqrt{5} + 10) = (10\sqrt{5})^2 - 10^2 = 500 - 100 = 400 \). Yes! Perfect, that's the area. So the width is \( 10(\sqrt{5} - 1) \) feet and length is \( 10(\sqrt{5} + 1) \) feet. But let's compute the numerical values. \( \sqrt{5} \approx 2.236 \), so width: \( 10(2.236 - 1) = 10(1.236) = 12.36 \) ft, length: \( 10(2.236 + 1) = 10(3.236) = 32.36 \) ft. Let's check area: \( 12.36 \times 32.36 \approx 400 \), which matches.

But maybe the problem expects an exact form or a simplified radical. Wait, let's re-express:

Width: \( -10 + 10\sqrt{5} = 10(\sqrt{5} - 1) \) ft

Length: \( 10 + 10\sqrt{5} = 10(\sqrt{5} + 1) \) ft

Alternatively, we can write them as \( 10\sqrt{5} - 10 \) and \( 10\sqrt{5} + 10 \), but factoring out 10 is neater.

Wait, but maybe I made a mistake in setting up the equation. Let's recheck:

Let width = \( w \), length = \( w + 20 \). Area = \( w(w + 20) = 400 \). So \( w^2 + 20w - 400 = 0 \). Correct. Quadratic formula: \( w = \frac{-20 \pm \sqrt{400 + 1600}}{2} = \frac{-20 \pm \sqrt{2000}}{2} = \frac{-20 \pm 10\sqrt{20}}{2} = \frac{-20 \pm 20\sqrt{5}}{2} = -10 \pm 10\sqrt{5} \). Positive solution: \( -10 + 10\sqrt{5} \), since \( 10\sqrt{5} \approx 22.36 \), so \( 22.36 - 10 = 12.36 \), which is positive. Then length is \( 12.36 + 20 = 32.36 \), which is \( 10\sqrt{5} + 10 \approx 22.36 + 10 = 32.36 \). Correct.

So the width is \( 10(\sqrt{5} - 1) \) feet (or \( -10 + 10\sqrt{5} \) feet) and the length is \( 10(\sqrt{5} + 1) \) feet (or \( 10 + 10\sqrt{5} \) feet).

But let's write them in simplified form:

Width: \( 10\sqrt{5} - 10 \) ft (or factor as \( 10(\sqrt{5} - 1) \) ft)

Length: \( 10\sqrt{5} + 10 \) ft (or factor as \( 10(\sqrt{5} + 1) \) ft)

Alternatively, if we solve the quadratic equation again, maybe there's a mistake in the setup? Wait, the problem says "length must be 20 feet longer than the width". So length = width + 20. Area = length * width = 400. So equation is correct: \( w(w + 20) = 400 \), leading to \( w^2 + 20w - 400 = 0 \). Correct.

So the dimensions are width \( 10(\sqrt{5} - 1) \) ft and length \( 10(\sqrt{5} + 1) \) ft. Let's compute the numerical values:

\( \sqrt{5} \approx 2.23607 \)

Width: \( 10(2.23607 - 1) = 10(1.23607) = 12.3607 \) ft ≈ 12.36 ft

Length: \( 10(2.23607 + 1) = 10(3.23607) = 32.3607 \) ft ≈ 32.36 ft

Checking area: 12.36 * 32.36 ≈ 400, which is correct.

So the width is \( 10\sqrt{5} - 10 \) feet (or \( 10(\sqrt{5} - 1) \) feet) and the length is \( 10\sqrt{5} + 10 \) feet (or \( 10(\sqrt{5} + 1) \) feet).

Answer:

The width of the pen is \( \boldsymbol{10(\sqrt{5} - 1)} \) feet (or \( \boldsymbol{10\sqrt{5} - 10} \) feet) and the length of the pen is \( \boldsymbol{10(\sqrt{5} + 1)} \) feet (or \( \boldsymbol{10\sqrt{5} + 10} \) feet).

(If numerical approximation is preferred, width ≈ 12.36 ft, length ≈ 32.36 ft, but the exact form is \( 10\sqrt{5} - 10 \) ft and \( 10\sqrt{5} + 10 \) ft.)