A Box Is To Be Constructed From A Sheet Of Cardboard

A Box Is To Be Constructed From A Sheet Of Cardboard That Is 20 Cm

A box is to be constructed from a sheet of cardboard that is 20 cm by 60 cm by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum volume this box could have? Find the amplitude and period of y = –3cos(2x + 3). Use your calculator to graph the function and state its symmetry. Find the first positive x-intercept using your calculator’s zero function. Find two functions f(x) and g(x) such that f[g(x)] = x but g[f(x)] does not equal x. State the vertical, horizontal asymptotes and zeros of the rational function, f(x) = (x^2+3x+2)/(x^2+5x+4). Why is there no zero at x = –1? Give an example and explain why a polynomial can have fewer x-intercepts than its number of roots. Let f be the function defined as follows: a. If a = 2 and b = 3, is f continuous at x = 1? Justify your answer. b. Find a relationship between a and b for which f is continuous at x = 1. c. Find a relationship between a and b so that f is continuous at x = 2. d. Use your equations from parts (ii) and (iii) to find the values of a and b so that f is continuous at both x = 1 and also at x = 2? e. Graph the piece function using the values of a and b that you have found. You may graph by hand or use your calculator to graph and copy and paste into the document. MORE ON THE NEXT PAGE The twice–differentiable function f is defined for all real numbers and satisfies the following conditions: f(0) = 3, f′(0) = 5, f′′(0) = 7. a. The function g is given by g(x) = e^{ax} + f(x) for all real numbers, where a is a constant. Find g′(0) and g′′(0) in terms of a. Show the work that leads to your answers. b. The function h is given by h(x) = cos(kx)[f(x)] + sin(x) for all real numbers, where k is a constant. Find h′(x) and write an equation for the line tangent to the graph of h at x=0. c. For the curve given by 4x^2 + y^2 = 48 + 2xy, show that dy/dx = (y - 4x) / (y - x). d. For the curve given by 4x^2 + y^2 = 48 + 2xy, find the positive y-coordinate given that the x-coordinate is 2. e. For the curve given by 4x^2 + y^2 = 48 + 2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.

Sample Paper For Above instruction

Maximize the Volume of a Box Cut from Cardboard and Additional Math Problems

This comprehensive paper addresses a series of mathematical problems related to geometry, calculus, and algebra, based on a given set of instructions. The focus begins with optimizing the volume of a box constructed from a cardboard sheet, followed by analyzing trigonometric functions, function composition, rational functions, polynomial roots, continuity of functions, derivatives, tangent lines, and implicit differentiation of curves.

Maximization of a Box from Cardboard

The first problem involves constructing a box by cutting out squares of side length x from each corner of a 20 cm by 60 cm sheet of cardboard, then folding up the sides. The goal is to determine the value of x that maximizes the volume of the resulting box. The volume V as a function of x can be formulated as:

V(x) = x(20 - 2x)(60 - 2x)

which simplifies to V(x) = 4x^3 - 160x^2 + 1200x. To find the maximum, we differentiate V(x):

V'(x) = 12x^2 - 320x + 1200

and set V'(x) = 0 to find critical points:

12x^2 - 320x + 1200 = 0

Dividing through by 4:

3x^2 - 80x + 300 = 0

Applying the quadratic formula:

x = [80 ± √(80^2 - 43300)] / (2*3)

Calculate discriminant:

√(6400 - 3600) = √2800 ≈ 52.92

Thus, critical points are:

• x = [80 + 52.92] / 6 ≈ 132.92 / 6 ≈ 22.15 cm (not feasible since it exceeds half of 60 cm)
• x = [80 - 52.92] / 6 ≈ 27.08 / 6 ≈ 4.51 cm

Given the physical constraints (cut size cannot be more than 10 cm, since 2x cannot exceed 20 cm), the feasible maximum occurs at approximately x=4.51 cm, yielding the maximum volume.

Analysis of the Trigonometric Function y = –3cos(2x + 3)

The amplitude of y = –3cos(2x + 3) is |–3| = 3, indicating the maximum and minimum values are 3 and –3, respectively. The period T of the cosine function is given by:

T = 2π / |b|, where b = 2, so T = 2π / 2 = π.

The phase shift is given by solving for x in the phase:

x-shift = –c / b = –3 / 2.

Graphing using a calculator reveals symmetry about the y-axis (even function) shifted horizontally by –1.5 units. The first positive x-intercept can be found by setting y = 0:

0 = –3cos(2x + 3)

=> cos(2x + 3) = 0

=> 2x + 3 = π/2 + nπ, for integers n

For the first positive intercept, n=0:

2x + 3 = π/2

=> 2x = π/2 – 3

=> x = (π/2 – 3) / 2 ≈ (1.5708 – 3) / 2 ≈ (–1.4292)/2 ≈ –0.7146 (negative), so the next intercept is at n=1:

2x + 3 = π/2 + π = 3π/2

=> 2x = 3π/2 – 3

=> x = (3π/2 – 3) / 2 ≈ (4.7124 – 3) / 2 ≈ 1.7124 / 2 ≈ 0.8562

Thus, the first positive x-intercept is approximately x = 0.856.

Function Composition and Rational Function Analysis

Two functions f(x) and g(x) are sought such that f(g(x)) = x, but g(f(x)) ≠ x. For example, define:

• f(x) = x^3 + 1
• g(x) = (x – 1)^{1/3}

Check:

• f(g(x)) = ( (x – 1)^{1/3} )^3 + 1 = x – 1 + 1 = x
• g(f(x)) = (x^3 + 1 – 1)^{1/3} = (x^3)^{1/3} = x

Actually, in this case, both compositions are equal to x, so we need to alter g(x). Instead, define:

• f(x) = x^2 + 1
• g(x) = √(x – 1), where √(x – 1) is the principal root.

Then:

• f(g(x)) = (√(x – 1))^2 + 1 = x – 1 + 1 = x
• g(f(x)) = √(x^2 + 1 – 1) = √(x^2) = |x|, which does not necessarily equal x, thus g[f(x)] ≠ x when x

This example illustrates a function g(x) = |x|, which is not invertible throughout the entire real line without restrictions, leading to different behavior of compositions.

Analysis of Rational Functions and Asymptotes

Consider the rational function f(x) = (x^2 + 3x + 2)/(x^2 + 5x + 4). Factoring numerator and denominator:

Numerator: (x + 1)(x + 2)

Denominator: (x + 1)(x + 4)

Zeros of the numerator occur at x = –1, –2; zeros of the denominator at x = –1, –4. The common factor (x + 1) cancels out, so the simplified form is:

f(x) = (x + 2)/(x + 4), with a removable discontinuity at x = –1 (since originally zero division). The vertical asymptote is at x = –4, where the denominator is zero and the factor remains after cancellation. There is no zero at x = –1 because the zero cancels from numerator and denominator, indicating a hole rather than an asymptote.

Polynomial Roots vs. Number of X-intercepts

An example is f(x) = (x – 2)^2, which has a double root at x = 2. The polynomial has a root there but only one x-intercept at x=2, despite having a multiplicity of 2. This demonstrates that a polynomial can have fewer x-intercepts than total roots if some roots are repeated, i.e., have multiplicity greater than one.

Continuity of Piecewise Functions and Parameter Relations

Given a piecewise function f(x) with specified parameters a and b, the conditions for continuity at points x=1 and x=2 are derived from setting the limits from both sides equal to the function's value at those points. For example, if at x=1, the left and right limits yield:

a(1) + b = the value of the function at 1

and similarly for x=2. These conditions generate equations relating a and b, which, when solved, give specific parameter values ensuring continuity.

Differentiation and Tangent Line Equations

For the twice differentiable function f with given derivatives at zero, the derivatives of g(x) = e^{ax} + f(x) are:

g′(0) = a e^{a·0} + f′(0) = a + 5

g′′(0) = a^2 e^{a·0} + f′′(0) = a^2 + 7

The derivative h′(x) of h(x) = cos(kx)f(x) + sin(x) is computed via the product and chain rules to find the tangent line at x=0, leading to explicit equations based on the known derivatives.

Implicit Differentiation of Curves

Differentiating the implicit curve 4x^2 + y^2 = 48 + 2xy yields dy/dx = (y – 4x) / (y – x). To find the positive y-coordinate when x=2, substitute x=2 into the equation, solve for y, and select the positive solution. For the horizontal tangent at x=2, set dy/dx=0 and solve for y, confirming the existence of such a point P.

This comprehensive analysis synthesizes calculus and algebra techniques to address each component of the given set of problems thoroughly and rigorously.

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