Homework 4: Maximum Or Minimum Name ID Question 1: What Is T

Homework 4 Maximum Or Minimum Name Idquestion 1what Is The M

Determine the maximum or minimum point for various curves and solve related optimization problems based on given functions and conditions. Analyze functions to find local maxima and minima using derivatives, and optimize geometric and physical scenarios such as area, volume, and height under specific constraints.

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Optimization problems are fundamental in calculus, providing insights into maximum efficiency, highest or lowest points, and optimal resource allocation. These problems often involve finding local maxima and minima of functions, which are critical points where the function’s derivative equals zero or does not exist. The following discussion explores methods for determining these points and applies them to real-world contexts, including geometric shapes, physical phenomena, and community health assessments.

Identifying maxima and minima begins with analyzing the first derivative of a function, \(f'(x)\). Critical points occur where \(f'(x) = 0\) or \(f'(x)\) is undefined. To classify these critical points, the second derivative, \(f''(x)\), is utilized. If \(f''(x) > 0\) at a critical point, the function has a local minimum there; if \(f''(x)

The second problem involves the quartic function \( y = 4x - x^4 \). Differentiation gives \( y' = 4 - 4x^3 \). Setting \( y' = 0 \) results in \( 4 - 4x^3 = 0 \), which simplifies to \( x^3 = 1 \), so \( x=1 \). Further, the second derivative \( y'' = -12x^2 \) is negative at \( x=1 \), indicating a local maximum. The function's value at this point is \( y = 4(1) - 1^4 = 4 - 1 = 3 \), so the maximum point is \((1, 3)\). This confirms choice D as the correct answer.

In the case of \( y = -3x^2 - 12x + 5 \), differentiation yields \( y' = -6x - 12 \). Setting \( y' =0 \) gives \( x = -2 \). The second derivative \( y'' = -6 \) is negative, indicating a local maximum at \( x=-2 \). Evaluating \( y \) at \( x=-2 \), we find \( y = -3(-2)^2 - 12(-2) + 5 = -3(4) + 24 + 5 = -12 + 24 + 5 = 17 \). Thus, the maximum point is \((-2, 17)\), matching choice B.

Optimization problems are also prevalent in geometry, such as maximizing area or volume under given constraints. For a rectangle with a perimeter of 20 meters, the area \( A = x \times y \), with \( y = (20 - 2x)/2 \). Simplifying, \( A = x(10 - x) \). This function reaches its maximum when \( x = 5 \), yielding \( y = 10 - 5= 5 \) and \( A= 25 \) m². Interestingly, the maximum area of a rectangle with a fixed perimeter occurs when the rectangle is a square, with area \( (20/4)^2 = 25 \) m², consistent with choice D.

The volume of a square prism with cross-section \( x \) units by \( x \) units and length \( (9 - 2x) \) units is \( V = x^2 (9 - 2x) \). Differentiating with respect to \( x \): \( V' = 2x (9 - 2x) + x^2 (-2) = 18x - 4x^2 - 2x^2 = 18x - 6x^2 \). Setting \( V' = 0 \) yields \( x (18 - 6x) = 0 \), so either \( x=0 \) or \( x=3 \). The latter gives maximum volume: \( V = 3^2 (9 - 6) = 9 \times 3= 27 \) units³, matching choice C.

In physical scenarios like projectile motion, maximizing height involves setting the derivative of height \( h = 5t(4 - t) \) to zero. Expanding \( h = 20t - 5t^2 \), the derivative \( h' = 20 - 10t \) equals zero at \( t=2 \). Substituting back, \( h=20(2) - 5(2)^2= 40 - 20= 20 \) meters. Thus, the maximum height is 20 meters, aligning with choice A.

Analyzing cubic functions for relative extrema involves setting the first derivative to zero. For \( f(x)= x^3 + 3x^2 - 24x \), \( f'(x)= 3x^2 + 6x - 24 \). Setting this to zero, \( 3x^2 + 6x - 24=0 \), simplifies to \( x^2 + 2x - 8=0 \). The solutions are \( x= -1 \pm \sqrt{1 + 8} = -1 \pm 3 \), giving \( x= 2 \) and \( x= -4 \). Evaluating \( f \) at these points indicates local maxima and minima, with the maximum value at \( x= 2 \), where, \( f(2)= 8 +12 -48= -28 \), which matches choice A for the maximum.

For \( f(x)= x^3(x+4) \), expanding gives \( f(x)= x^4 + 4x^3 \). Differentiating yields \( f'(x)= 4x^3 + 12x^2= 4x^2 (x + 3) \). Setting to zero, \( x= 0 \) or \( x= -3 \). To identify extrema, the second derivative \( f''(x)=12x^2 + 24x \) is used. At \( x= -3 \), \( f''= 12(9) +24(-3)=108-72=36>0 \), indicating a local minimum there and a local maximum at \( x=0 \), where \( f(0)= 0 \). The function's maximum value is thus 0, corresponding to selection B.

Analyzing the cubic function \( f(x)= 2x^3 - 24x - 32 \), its derivative is \( f'(x)= 6x^2 - 24 \), which equals zero at \( x= \pm 2 \). The second derivative is \( f''(x)= 12x \). At \( x= 2 \), \( f''=24 > 0 \), indicating a local minimum, whereas at \( x= -2 \), \( f'' = -24

The minimization of \( x^2 + y^2 \) under the linear constraint \( 2x + y=10 \) can be approached using substitution or Lagrange multipliers. Substituting \( y= 10 - 2x \), the expression becomes \( x^2 + (10 - 2x)^2 \). Simplify to \( x^2 + 100 - 40x + 4x^2= 5x^2 - 40x + 100 \). The minimum occurs where its derivative equals zero: \( 10x - 40= 0 \Rightarrow x= 4 \). Correspondingly, \( y= 10-8= 2 \). Substituting back, the minimum value is \( 4^2 + 2^2= 16+4=20 \), which matches choice B.

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