A rancher has 200 feet of fencing with which to enclose two adjacent rectangular corrals (see figure). What dimensions should be used so that the enclosed area will be a maximum?
The Correct Answer and Explanation is :
To maximize the area of two adjacent rectangular corrals enclosed by 200 feet of fencing, let’s determine the optimal dimensions.
Step 1: Variables and Relationships
Let:
- xx = length of each corral (in feet),
- yy = width of each corral (in feet).
The total perimeter is made of:
- Two lengths (2x2x),
- Three widths (3y3y).
The fencing constraint is: 2x+3y=200.2x + 3y = 200.
The area AA of the corrals is: A=x⋅y.A = x \cdot y.
Step 2: Solve for yy in terms of xx
From 2x+3y=2002x + 3y = 200: y=200−2×3.y = \frac{200 – 2x}{3}.
Step 3: Express Area as a Function of xx
Substitute y=200−2x3y = \frac{200 – 2x}{3} into A=x⋅yA = x \cdot y: A(x)=x⋅200−2×3.A(x) = x \cdot \frac{200 – 2x}{3}. A(x)=200x−2×23.A(x) = \frac{200x – 2x^2}{3}.
Simplify: A(x)=13(200x−2×2).A(x) = \frac{1}{3} \left( 200x – 2x^2 \right).
Step 4: Maximize A(x)A(x)
Find the critical points of A(x)A(x) by taking the derivative: A′(x)=13(200−4x).A'(x) = \frac{1}{3} (200 – 4x).
Set A′(x)=0A'(x) = 0: 200−4x=0⇒x=50.200 – 4x = 0 \quad \Rightarrow \quad x = 50.
Step 5: Verify Maximum and Solve for yy
When x=50x = 50: y=200−2(50)3=200−1003=1003≈33.33 feet.y = \frac{200 – 2(50)}{3} = \frac{200 – 100}{3} = \frac{100}{3} \approx 33.33 \, \text{feet}.
The maximum area occurs when: Length (x)=50 feet,Width (y)=33.33 feet.\text{Length (x)} = 50 \, \text{feet}, \quad \text{Width (y)} = 33.33 \, \text{feet}.
Explanation:
The total fencing of 200 feet creates a balance between length and width, ensuring both are large enough to maximize the area. By differentiating the area function and solving for the critical points, we identify the dimensions that optimize the enclosed area. The result aligns with the principle of efficiently distributing resources for maximum utility.