Math Problem Statement
Solution
Let's break down the problem and solution based on the image you uploaded.
Problem:
A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals. We are tasked with:
- Writing the area, , of the corral as a function of .
- Finding the values of and that will maximize the area.
(a) Write the area of the corral as a function of :
From the problem, we know that:
- The perimeter of the corral is constrained by the 200 feet of fencing.
- The total perimeter includes four vertical segments of length and three horizontal segments of length .
The perimeter constraint is: 4y + 3x = 200 \quad \text{(since there are 4 vertical segments of height yx)} We can solve for :
Now, the area of the corral, , is the product of length and width:
Simplifying this: Thus, the area function is:
(b) Find the values of and that will produce the maximum area:
To maximize the area, we take the derivative of with respect to and set it equal to zero to find the critical points.
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Derivative of :
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Set the derivative equal to zero: Solving for :
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Now, substitute back into the expression for :
So, the dimensions that maximize the area are approximately:
These values will give the maximum possible area.
Do you have any questions or want more details?
Here are five follow-up questions for practice:
- How would the area function change if there were three vertical segments and four horizontal segments?
- What would happen to the optimal dimensions if the total fencing were increased to 300 feet?
- How do you find the maximum area without using calculus?
- How does the shape of the corral affect the maximum area that can be enclosed?
- What is the importance of the derivative in finding maximum and minimum values?
Tip: Whenever you have a perimeter constraint problem, always express one variable in terms of the other using the given constraints to simplify the area equation.
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Math Problem Analysis
Mathematical Concepts
Quadratic Functions
Optimization
Derivatives
Formulas
Perimeter constraint: 4y + 3x = 200
Area function: A(x) = 50x - (3x^2)/4
Derivative of A(x): A'(x) = 50 - (3/2)x
Quadratic equation solution: x = (-b ± sqrt(b^2 - 4ac)) / 2a
Theorems
Quadratic Formula
First Derivative Test for Maximum
Suitable Grade Level
Grade 11-12
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