Ch. 4 - Applications of Derivatives

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

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.5.35

Chapter 4, Problem 4.5.35

35. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.

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Identify the right triangle with legs of lengths 3 and 4, and hypotenuse 5. The rectangle is inscribed such that one vertex is at the right angle, and the opposite vertex is on the hypotenuse.

Let the width of the rectangle be 'w' and the height be 'h'. The area of the rectangle is given by A = w * h.

Use similar triangles to express 'h' in terms of 'w'. The triangle formed by the hypotenuse and the rectangle is similar to the original triangle, so the ratio of corresponding sides is equal: h/3 = (4-w)/4.

Solve the equation from the similar triangles to express 'h' in terms of 'w': h = 3(4-w)/4.

Substitute the expression for 'h' into the area formula: A = w * (3(4-w)/4). Differentiate A with respect to 'w', set the derivative equal to zero, and solve for 'w' to find the dimensions that maximize the area.

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Key Concepts

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

Optimization

Optimization involves finding the maximum or minimum value of a function within a given set of constraints. In this problem, the goal is to maximize the area of the rectangle inscribed in the right triangle. This requires setting up an equation for the area in terms of the rectangle's dimensions and using calculus techniques to find the maximum value.

Similar Triangles

Similar triangles have proportional sides and identical angles. In this problem, the inscribed rectangle creates two smaller triangles within the right triangle. By using the properties of similar triangles, we can express the dimensions of the rectangle in terms of the triangle's sides, which is crucial for setting up the optimization problem.

Derivative Test

The derivative test is used to determine the local maxima and minima of a function. After expressing the area of the rectangle as a function of one variable, taking its derivative and setting it to zero helps find critical points. Evaluating the second derivative at these points confirms whether they correspond to a maximum or minimum area.

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The Second Derivative Test: Finding Local Extrema

Related Practice

Textbook Question

Finding Indefinite Integrals

In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

∫(1/x² − x² − 1/3) dx

Textbook Question

Finding Critical Points

In Exercises 41–50, determine all critical points and all domain endpoints for each function.

f(x) = x(4 − x)³

Textbook Question

Estimate the open intervals on which the function y = ƒ(𝓍) is

a. increasing.

b. decreasing.

c. Use the given graph of ƒ' to indicate where any local extreme

values of the function occur, and whether each extreme

is a relative maximum or minimum.

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Textbook Question

Roots (Zeros)

Show that the functions in Exercises 19–26 have exactly one zero in the given interval.

r(θ) = 2θ − cos²θ + √2, (−∞, ∞)

Textbook Question

93. The accompanying figure shows a portion of the graph of a twice-differentiable function y=f(x). At each of the five labeled points, classify y' and \(\y\)'' as positive, negative, or zero.

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Textbook Question

32. Answer Exercise 31 if one piece is bent into a square and the other into a circle.