Mr. Honda Wants To Find The Length And Width That Would Maximize The Area Enclosed By The Fence. He Is Completing The Square As Shown Below, But He Still Needs To Find The Value Of $c$.Given:${ A = -x^2 + 50x }$[ A = -(x -

Introduction

In mathematics, optimization problems often involve finding the maximum or minimum value of a function. In this article, we will explore a classic optimization problem involving a fence and a square. Mr. Honda wants to find the length and width of a square that would maximize the area enclosed by the fence. To solve this problem, we will use the concept of completing the square and algebraic manipulation.

The Problem

Given the function:

$$A = -x^2 + 50x$$

We need to find the value of $c$ that would maximize the area enclosed by the fence. To do this, we will first complete the square and then use algebraic manipulation to find the maximum value of the function.

Completing the Square

Completing the square is a technique used to rewrite a quadratic function in the form $(x - h)^2 + k$. This form is useful for finding the maximum or minimum value of a quadratic function. To complete the square, we need to add and subtract a constant term to the function.

Let's start by adding and subtracting $(50/2)^2 = 625$ to the function:

$$A = -x^2 + 50x + 625 - 625$$

Now, we can rewrite the function as:

$$A = -(x - 25)^2 + 625$$

Finding the Maximum Value

Now that we have completed the square, we can see that the function is in the form $(x - h)^2 + k$. The maximum value of the function occurs when the squared term is equal to zero, which means that $x - h = 0$. In this case, $h = 25$, so the maximum value occurs when $x = 25$.

To find the maximum value of the function, we can substitute $x = 25$ into the function:

$$A = -(25 - 25)^2 + 625$$

$$A = 0 + 625$$

$$A = 625$$

Conclusion

In this article, we have used the concept of completing the square to find the maximum value of a quadratic function. We have shown that the maximum value of the function occurs when $x = 25$, and the maximum value is equal to $625$. This result can be used to find the length and width of a square that would maximize the area enclosed by a fence.

Maximizing the Area Enclosed by a Fence

Now that we have found the maximum value of the function, we can use this result to find the length and width of a square that would maximize the area enclosed by a fence. Let's assume that the length of the square is $x$ and the width is $y$. The area of the square is given by:

$$A = xy$$

We want to find the values of $x$ and $y$ that would maximize the area enclosed by the fence. To do this, we can use the result we found earlier, which is that the maximum value of the function occurs when $x = 25$.

Since the length of the square is $x$, we can set $x = 25$ to find the length of the square:

$$x = 25$$

Now, we need to find the width of the square. To do this, we can use the fact that the area of the square is given by:

$$A = xy$$

We know that the maximum value of the function occurs when $x = 25$, so we can substitute $x = 25$ into the equation:

$$A = 25y$$

We also know that the maximum value of the function is equal to $625$, so we can substitute $A = 625$ into the equation:

$$625 = 25y$$

Now, we can solve for $y$:

$$y = 625/25$$

$$y = 25$$

Conclusion

In this article, we have used the concept of completing the square to find the maximum value of a quadratic function. We have shown that the maximum value of the function occurs when $x = 25$, and the maximum value is equal to $625$. We have also used this result to find the length and width of a square that would maximize the area enclosed by a fence. The length of the square is $25$, and the width is also $25$.

Mathematical Formulas

• $A = -x^2 + 50x$
• $A = -(x - 25)^2 + 625$
• $A = xy$
• $x = 25$
• $y = 625/25$

References

• [1] "Completing the Square" by Math Open Reference
• [2] "Quadratic Functions" by Khan Academy

Glossary

• Completing the Square: A technique used to rewrite a quadratic function in the form $(x - h)^2 + k$.
• Quadratic Function: A function of the form $f(x) = ax^2 + bx + c$.
• Maximum Value: The largest value of a function.
• Minimum Value: The smallest value of a function.
  Q&A: Maximizing the Area Enclosed by a Fence =============================================

Frequently Asked Questions

In this article, we will answer some of the most frequently asked questions about maximizing the area enclosed by a fence.

Q: What is the main goal of maximizing the area enclosed by a fence?

A: The main goal of maximizing the area enclosed by a fence is to find the length and width of a square that would maximize the area enclosed by the fence.

Q: How do we find the maximum value of the function?

A: To find the maximum value of the function, we need to complete the square and then use algebraic manipulation to find the maximum value of the function.

Q: What is the formula for completing the square?

A: The formula for completing the square is:

$$A = -(x - h)^2 + k$$

Q: How do we find the value of h in the formula?

A: To find the value of h in the formula, we need to add and subtract $(b/2)^2$ to the function, where b is the coefficient of the x term.

Q: What is the maximum value of the function?

A: The maximum value of the function is equal to $k$ in the formula.

Q: How do we find the length and width of the square that would maximize the area enclosed by the fence?

A: To find the length and width of the square that would maximize the area enclosed by the fence, we need to use the result we found earlier, which is that the maximum value of the function occurs when $x = 25$. We can then set $x = 25$ to find the length of the square, and use the fact that the area of the square is given by $A = xy$ to find the width of the square.

Q: What is the length and width of the square that would maximize the area enclosed by the fence?

A: The length of the square is $25$, and the width is also $25$.

Q: What are some real-world applications of maximizing the area enclosed by a fence?

A: Some real-world applications of maximizing the area enclosed by a fence include:

• Designing a garden or a park with a maximum area
• Building a fence around a property with a maximum area
• Designing a storage facility with a maximum area

Q: How do we use calculus to maximize the area enclosed by a fence?

A: To use calculus to maximize the area enclosed by a fence, we need to find the derivative of the function and set it equal to zero to find the critical points. We can then use the second derivative test to determine whether the critical points are maximum or minimum values.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine whether a critical point is a maximum or minimum value. If the second derivative is positive at a critical point, then the critical point is a minimum value. If the second derivative is negative at a critical point, then the critical point is a maximum value.

Q: How do we use the second derivative test to determine whether a critical point is a maximum or minimum value?

A: To use the second derivative test to determine whether a critical point is a maximum or minimum value, we need to find the second derivative of the function and evaluate it at the critical point. If the second derivative is positive at the critical point, then the critical point is a minimum value. If the second derivative is negative at the critical point, then the critical point is a maximum value.

Conclusion

In this article, we have answered some of the most frequently asked questions about maximizing the area enclosed by a fence. We have discussed the main goal of maximizing the area enclosed by a fence, how to find the maximum value of the function, and how to use calculus to maximize the area enclosed by a fence. We have also discussed some real-world applications of maximizing the area enclosed by a fence and how to use the second derivative test to determine whether a critical point is a maximum or minimum value.