Math Problem Statement

You’re building five identical pens adjacent to each other with a total area of 1000 m². What dimensions should you use to minimize the amount of fence?

Solution

The problem you're solving involves finding the dimensions of five identical pens arranged side by side, where the total area is 1000m21000 \, m^2, and we need to minimize the amount of fencing used. Here's how we approach it:

1. Area Equation:

The total area of the five pens is given by: 1000=5xy1000 = 5xy where xx is the length of the pens and yy is the width of each pen. This equation can be rewritten to solve for xx: x=10005y=200yx = \frac{1000}{5y} = \frac{200}{y}

2. Fence Equation:

The total length of fencing required (perimeter) is given by: F=6x+10yF = 6x + 10y This includes the lengths of the horizontal sections (6 xx, as there are six vertical boundaries) and the vertical sections (10 yy, as there are five pens requiring five boundaries and two external ones).

3. Substituting xx into the Fence Equation:

Now substitute x=200yx = \frac{200}{y} into the equation for the fence: F=6(200y)+10y=1200y+10yF = 6 \left( \frac{200}{y} \right) + 10y = \frac{1200}{y} + 10y

4. Minimizing the Fence Equation:

To minimize the amount of fence used, take the derivative of FF with respect to yy and set it equal to zero: dFdy=1200y2+10\frac{dF}{dy} = -\frac{1200}{y^2} + 10 Setting this equal to zero: 1200y2+10=0-\frac{1200}{y^2} + 10 = 0 1200y2=10\frac{1200}{y^2} = 10 y2=120010=120y^2 = \frac{1200}{10} = 120 y=12010.95my = \sqrt{120} \approx 10.95 \, m

5. Finding xx:

Now, substitute y10.95y \approx 10.95 back into the equation for xx: x=200y=20010.9518.27mx = \frac{200}{y} = \frac{200}{10.95} \approx 18.27 \, m

Final Dimensions:

  • Width y10.95my \approx 10.95 \, m
  • Length x18.27mx \approx 18.27 \, m

These dimensions will minimize the amount of fencing used while keeping the total area of the pens at 1000m21000 \, m^2.

Would you like more details or have any questions?

Related Questions:

  1. How do we confirm that the calculated dimensions give the correct total area of 1000m21000 \, m^2?
  2. Can we use Lagrange multipliers to solve this optimization problem?
  3. What happens to the fence equation if the number of pens changes from 5 to another value?
  4. What if the area constraint changes to something other than 1000m21000 \, m^2?
  5. How does the shape of the pens (square vs rectangular) affect the fencing requirements?

Tip:

When optimizing a quantity, always check if the second derivative confirms a minimum or maximum!

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Math Problem Analysis

Mathematical Concepts

Optimization
Calculus
Geometry

Formulas

Area: 1000 = 5xy
Fence Equation: F = 6x + 10y
Minimization of F: dF/dy = -1200/y^2 + 10

Theorems

Optimization using derivatives

Suitable Grade Level

Grades 11-12, College level