Let's go step by step.

Step 1: Understanding the problem

• We have 42 yards of fencing.
• One side of the rectangular enclosure is along a wall, so no fencing is needed there.
• The fencing is used for the other three sides: two vertical sides and one horizontal side.
• The enclosure is divided into two equal pens, so we need an extra fence inside to divide them.

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Step 2: Define Variables

• Let x be the length of the rectangle perpendicular to the wall.
• Let y be the length parallel to the wall.

Since we have two vertical fences (both of length x) and two internal dividers (both of length x), plus one long fence of length y, the total fencing equation is:

$2x + x + y = 42$

which simplifies to:

$3x + y = 42$

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Step 3: Express y in terms of x

Solving for $y$:

$y = 42 - 3x$

This is the expression for the other dimension.

Answer for part (a):

$y = 42 - 3x \quad \text{yards}$

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Step 4: Expression for Area

$A = x \cdot y$

Substituting $y = 42 - 3x$:

$A = x(42 - 3x)$

$A = 42x - 3x^2$

Answer for part (b):

$A = 42x - 3x^2 \quad \text{square yards}$

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Step 5: Find Maximum Area

To maximize the area, take the derivative and set it to zero:

$\frac{dA}{dx} = 42 - 6x = 0$

Solving for $x$:

$6x = 42$

$x = 7$

Find $y$:

$y = 42 - 3(7) = 21$

Answer for part (c):

$7,21$

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Step 6: Maximum Area

Substituting $x = 7$ and $y = 21$:

$A_{\max} = 7 \times 21 = 147$

Answer for part (d):

$147 \quad \text{square yards}$

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Final Answers

a) $y = 42 - 3x$ yards
b) $A = 42x - 3x^2$ square yards
c) $7, 21$ yards
d) $147$ square yards

Would you like a graph or further explanation?

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Related Questions

1. How would the solution change if the wall were removed?
2. What happens if we had three pens instead of two?
3. Can we maximize the area using calculus only?
4. What are the second derivative and what do they tell us?
5. What happens to the maximum area if the available fencing changes?

Tip:

Always check units when solving word problems. A common mistake is missing a square or linear unit!