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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.5.6b
Chapter 4, Problem 4.5.6b

Suppose S = x + 2y is an objective function subject to the constraint xy = 50, for x > 0 and y > 0.
b. Find the absolute minimum value of S subject to the given constraint.

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1
First, express the constraint equation xy = 50 in terms of one variable. Solve for y in terms of x: y = 50/x.
Substitute y = 50/x into the objective function S = x + 2y to express S in terms of x alone: S = x + 2(50/x).
Simplify the expression for S: S = x + 100/x.
To find the critical points, take the derivative of S with respect to x. Use the power rule and the derivative of x^(-1) to find dS/dx.
Set the derivative dS/dx equal to zero and solve for x to find the critical points. Check these points to determine the absolute minimum value of S.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Objective Function

An objective function is a mathematical expression that defines a quantity to be maximized or minimized. In this case, S = x + 2y is the objective function, which we aim to minimize while adhering to certain constraints. Understanding how to manipulate and evaluate this function is crucial for finding optimal solutions.
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Constraints

Constraints are conditions that the variables in an optimization problem must satisfy. Here, the constraint xy = 50 restricts the values of x and y, ensuring they remain positive. Recognizing how constraints affect the feasible region is essential for determining the minimum value of the objective function.
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Lagrange Multipliers

Lagrange multipliers are a method used in optimization to find the local maxima and minima of a function subject to equality constraints. This technique involves introducing a new variable (the multiplier) to incorporate the constraint into the optimization process. Applying this method will help in finding the absolute minimum value of S under the given constraint.
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Related Practice
Textbook Question

{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>

a. Find the time at which the object first passes the rest position, y = 0. 

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Textbook Question

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b. A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100 m² (see figure). What are the dimensions of each pen that minimize the amount of fence that must be used? <IMAGE>

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{Use of Tech} Basketball shot A basketball is shot with an initial velocity of v ft/s at an angle of 45° to the floor. The center of the basketball is 8 ft above the floor at a horizontal distance of 18 feet from the center of the basketball hoop when it is released. The height h (in feet) of the center of the basketball after it has traveled a horizontal distance of x feet is modeled by the function h(x) = 32x² / v² + x + 8 (see figure). <IMAGE>



b. During the flight of the basketball, show that the distance s from the center of the basketball to the front of the hoop is s = √ (x - 17.25)² + ( -(4x² / 81) + x - 2)² (Hint: The diameter of the basketball hoop is 18 inches.) 

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Textbook Question

Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.


b. Does either g or h have a local extreme value at x = 2? Explain.

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Textbook Question

{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>

b. Find the time and the displacement when the object reaches its lowest point.

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Textbook Question

Two poles of heights m and n are separated by a horizontal distance d. A rope is stretched from the top of one pole to the ground and then to the top of the other pole. Show that the configuration that requires the least amount of rope occurs when Θ₁ = Θ₂ (see figure). <IMAGE>

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