Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.5.6

Chapter 4, Problem 4.5.6

6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

Verified step by step guidance

First, understand that the line segment forms a right triangle with the x-axis and y-axis. The vertices of the triangle are (0,0), (a,0), and (0,b).

The length of the line segment is given as 20 units. Using the distance formula, we have \( \sqrt{a^2 + b^2} = 20 \).

The area \( A \) of the triangle can be expressed as \( A = \frac{1}{2}ab \).

To find the maximum area, express \( b \) in terms of \( a \) using the equation from the distance formula: \( b = \sqrt{400 - a^2} \). Substitute this into the area formula to get \( A(a) = \frac{1}{2}a\sqrt{400 - a^2} \).

Differentiate \( A(a) \) with respect to \( a \) and set the derivative equal to zero to find the critical points. Solve for \( a \) and verify that the area is maximized when \( a = b \).

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Key Concepts

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

Equation of a Line

The equation of a line in the coordinate plane can be expressed in the form y = mx + c, where m is the slope and c is the y-intercept. For a line segment from (a, 0) to (0, b), the slope is -b/a, and the equation becomes y = (-b/a)x + b. Understanding this helps in determining the line's position and its intersection with the axes.

Area of a Triangle

The area of a triangle can be calculated using the formula (1/2) * base * height. In this context, the base and height are the x and y intercepts of the line segment, which are a and b, respectively. Thus, the area of the triangle formed is (1/2) * a * b, which is crucial for determining when this area is maximized.

Optimization

Optimization involves finding the maximum or minimum value of a function. Here, we need to maximize the area of the triangle, A = (1/2) * a * b, subject to the constraint that the line segment's length is 20 units, which gives the equation a^2 + b^2 = 400. Using calculus, particularly the method of Lagrange multipliers or substitution, helps find the optimal values of a and b.

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Related Practice

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Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

69. y' = x(x - 3)²

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Solve the initial value problems in Exercises 71–90.

dy/dx = 2x − 7, y(2) = 0

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Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.

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Exercises 45–48 give the acceleration a=d²s/dt², initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t.

a = 9.8, v(0) = −3, s(0) = 0

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[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.

f(x) = |x − 2| + |x + 3|, −5 ≤ x ≤ 5

Textbook Question

Initial Value Problems

Find the curve y = f(x) in the xy-plane that passes through the point (9,4) and whose slope at each point is 3√x.