In The Discussion Area Solve Your Two Custom Applications ✓ Solved

In The Discussion Area Solve Your Two Custom Application Problems Usi

In the discussion area, solve your two custom application problems using the concepts you have learned. Apply your CINs in the bracketed spaces (the first digit is [a], the second digit is [b], and so forth.)

Problems 1. A rock is thrown upward from the edge of a cliff. The rock follows the equation below. What is the greatest height above the ground that this rock will go?

2. You are designing a rectangular enclosure with [a] rectangular interior sections separated by parallel walls. If you have 300* [c] feet of fencing, what is the maximum area that can be enclosed? Use the following numbers for the alphabets: a=2, b=9, c=7, d=8, e=3

Sample Paper For Above instruction

Introduction

The application of mathematical concepts such as quadratic functions and optimization techniques is fundamental in solving real-world problems. This paper addresses two such problems: first, determining the maximum height reached by a projectile, and second, maximizing the area of an enclosure under fencing constraints. Both problems employ algebraic and calculus-based methods to find optimal solutions, illustrating the practical utility of mathematical analysis.

Problem 1: Maximum Height of a Thrown Rock

The first problem involves a rock thrown upward from a cliff, described by a quadratic equation typically of the form:

h(t) = -at^2 + bt + c

where h(t) is the height at time t, and the coefficients are known constants derived from initial velocity and gravity. Using the given variables, with a = 2 and b = 9 (from the provided numeric substitutions), the equation becomes:

h(t) = -2t^2 + 9t + c

The maximum height corresponds to the vertex of the parabola, which can be found using the vertex formula t = -b/(2a). Substituting the values:

t = -9/(2 * -2) = -9/(-4) = 2.25 seconds.

Next, compute the height at t = 2.25 seconds:

h(2.25) = -2(2.25)^2 + 9(2.25) + c

Assuming c is zero (the height at the initial point), we have:

h(2.25) = -2(5.0625) + 20.25 = -10.125 + 20.25 = 10.125 feet.

This is the greatest height the rock will reach above the ground.

Problem 2: Maximizing Area of a Rectangular Enclosure

For the second problem, with [a] interior sections separated by parallel walls, and a total fencing length of 300 * [c] feet, we need to determine the maximum enclosed area.

Given the values: a=2, c=7, so total fencing is:

F = 300 * 7 = 2100 feet.

Suppose the rectangle has a length L and width W. The total fencing used includes the perimeter of the rectangle plus the additional walls separating the sections:

F = 2L + (a - 1) * W + 2W

However, considering each interior section separated by walls parallel to the width, and assuming the walls are included within the fencing, the total fencing can be expressed as:

F = 2L + (a * W)

Given the fencing constraint, we can set:

2L + a * W = 2100

Where the goal is to maximize the area A = L * W.

Express L in terms of W:

L = (2100 - a * W)/2

Substitute into the area function:

A(W) = W (2100 - a W)/2 = (2100 W - a W^2)/2

With a = 2:

A(W) = (2100W - 2 W^2)/2 = 1050W - W^2

To find the maximum, take the derivative of A(W) with respect to W:

dA/dW = 1050 - 2W

Set derivative to zero to find critical points:

1050 - 2W = 0 => W = 525

Calculate L at W = 525:

L = (2100 - 2*525)/2 = (2100 - 1050)/2 = 1050/2 = 525

Therefore, the maximum enclosed area is:

A = L W = 525 525 = 275,625 square feet.

Conclusion

Applying quadratic and optimization principles, we determined that the maximum height of the projectile is approximately 10.125 feet, achieved at 2.25 seconds after launch. For the enclosure, the optimal design involves a square with sides of 525 feet,ensuring maximum area within the fencing constraints. These solutions exemplify how mathematical concepts facilitate efficient problem-solving in practical scenarios.

References

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