Calculus lies at the heart of modern mathematics, serving as a powerful tool for analyzing change and motion. From determining the slope of a curve to calculating the area under a complex function, its applications are vast and far-reaching. However, mastering calculus requires more than just memorizing formulas – it demands a deep understanding of its principles and the ability to apply them creatively to solve diverse problems.

In this blog post, we'll tackle two master-level calculus questions, showcasing the expertise of our Calculus Assignment Solver. Through detailed explanations and step-by-step solutions, you'll gain valuable insights into the thought processes behind solving complex calculus problems. So, let's dive in and explore the fascinating world of calculus together!

Question 1: Finding the Limit

Consider the function f(x) = (3x2 - 2x + 5) / (2x - 1). Determine the limit of f(x) as x approaches 1.

Solution:

To find the limit of f(x) as x approaches 1, we can't directly substitute x = 1 into the function since it results in division by zero. Instead, we'll use algebraic manipulation to simplify the expression and then evaluate the limit.

First, let's factorize the numerator:
f(x) = (3x2 - 2x + 5) / (2x - 1)
= [(3x - 1)(x - 5)] / (2x - 1)

Now, we can cancel out the common factor of (x - 1) from both the numerator and the denominator:
f(x) = (3x - 1)(x - 5) / (2x - 1)
= (3x - 1)(x - 5) / (2x - 1)(x - 1)

As x approaches 1, the term (x - 1) in the denominator approaches zero, while the other terms remain finite. Thus, we can evaluate the limit by directly substituting x = 1 into the simplified expression:
lim (x->1) f(x) = (3(1) - 1)(1 - 5) / (2(1) - 1)(1 - 1)
= (3 - 1)(-4) / (2 - 1)(0)
= (-2)(-4) / (1)(0)
= 8 / 0

Since the denominator approaches zero while the numerator remains finite, the limit is undefined (approaches infinity).

Question 2: Optimization Problem

A rectangular box with an open top is to be constructed from a rectangular piece of cardboard measuring 16 inches by 20 inches. Determine the dimensions of the box that will maximize its volume.

Solution:

Let's denote the length, width, and height of the box as l, w, and h, respectively. Since the box has an open top, its volume V is given by V = lwh.

We're given that the cardboard piece measures 16 inches by 20 inches, and we need to determine the dimensions of the box that will maximize its volume. Since the box is constructed from the cardboard piece, the dimensions of the box must satisfy the constraint imposed by the cardboard.

The constraint equation is given by the perimeter of the base of the box, which should be equal to the perimeter of the cardboard piece:
2l + 2w = 16 + 20
2l + 2w = 36
l + w = 18
l = 18 - w

Now, we can express the volume V in terms of a single variable, w, by substituting l = 18 - w into the volume formula:
V(w) = (18 - w)wh
= 18w - w2

To find the maximum volume, we'll take the derivative of V(w) with respect to w and set it equal to zero:
V'(w) = dV/dw = 18 - 2w

Setting V'(w) equal to zero and solving for w, we find the critical point:
18 - 2w = 0
2w = 18
w = 9

Now, we need to check the endpoints of the interval [0, 18] and the critical point w = 9 to determine which yields the maximum volume.

When w = 0, l = 18, and when w = 18, l = 0. These values correspond to degenerate cases where the box reduces to a flat sheet, resulting in a volume of zero.

Thus, the maximum volume occurs at w = 9 inches. Substituting w = 9 into the constraint equation, we find l = 18 - 9 = 9 inches.

Therefore, the dimensions of the box that maximize its volume are 9 inches by 9 inches by h inches, where h can vary.

By dissecting these master-level calculus questions, we've not only provided solutions but also illuminated the strategic thinking required to tackle such problems effectively. At MathsAssignmentHelp.com, we empower students to excel in calculus and beyond, offering personalized assistance tailored to their learning objectives. Whether you're grappling with limits, derivatives, or optimization problems, our team of experts is here to guide you every step of the way. Let's embark on a journey of mathematical discovery together – because with the right support, there's no limit to what you can achieve!