Solving a Complex Additional Mathematics Problem: A Step-by-Step Guide
When tackling complex mathematics problems, especially those in Additional Mathematics, it's often the case that the problem statement is not well-defined. This can make solving the problem quite challenging. In this article, we'll walk through the solution to one such challenging problem, highlighting the step-by-step approach to finding the dimensions that maximize the volume of a combined shape.
Understanding the Problem Statement
The given problem involves a shape composed of a rectangular portion and a semicircular portion. The total volume of the shape is given by:
1. Volume Expression
The volume of the piece of the shape is the sum of the volume of the rectangular portion:
1 middot; y middot; 2x
and the volume of the semicircular portion:
1/2 middot; π middot; x^2.
Therefore, the total volume V is:
V 2x 1/2 middot; π middot; x^2
Perimeter Constraint
A key constraint for the problem is the perimeter of the combined shape, which includes the rectangle and the semicircle. This perimeter is given as:
60 2y 2x π middot; x / 2.
By simplifying the perimeter equation:
60 2x 2y πx/2.
Rearranging this equation to solve for y:
y 30 - (π/2) middot; x.
Substituting y into the volume equation:
V 2x middot; [30 - (π/2) middot; x] 1/2 middot; π middot; x^2.
Expanding and simplifying:
V 6 - πx^2 1/2 middot; πx^2.
V 6 - πx^2/2.
V πx^2 (30 - π/2x).
Optimizing the Volume
To find the dimensions that maximize the volume, we need to take the derivative of V with respect to x and set it equal to zero.
2. Derivative and Optimizing
V πx^2 (30 - π/2x).
Taking the derivative with respect to x:
dV/dx πx (30 - π/2x) πx^2 middot; (-π/2).
Setting the derivative equal to zero to find the critical points:
πx (30 - π/2x) - π^2/2 x^2 0.
πx (30 - π/2x - π/2x) 0.
30 - πx 0.
Solving for x:
x 60 / (π middot; 2).
x 60 / (2π).
Finding the Maximum Volume
Once we have the value of x, we can substitute it back into the volume formula to find the maximum volume:
V πx^2 (30 - π/2x).
V π (60/2π)^2 (30 - π/2 middot; 60/2π).
V π (60/2π)^2 (30 - 30/2).
V π (60/2π)^2 (15).
V π (30/π)^2 (15).
V π (900/π^2) (15).
V (900/π) (15).
3. Final Calculation
V (13500/π) / π.
V 13500/π^2.
V ≈ 252.0446.
When the volume is maximized, y x.
The final shape is a square with two of its corners rounded off, forming the semicircle.
Conclusion
Solving complex Additional Mathematics problems requires careful attention to detail and a step-by-step approach. Understanding the constraints and expressing the volume in terms of a single variable is crucial for solving the problem. This guide should help you tackle similar problems in your studies.