Ch. 4 - Applications of Derivatives

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

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.5.41

Chapter 4, Problem 4.5.41

41. Among all triangles in the first quadrant formed by the x-axis, the y-axis, and tangent lines to the graph of y=3x-x^2, what is the smallest possible area?

Verified step by step guidance

Identify the function y = 3x - x^2 and note that the problem involves finding the area of triangles formed by tangent lines to this curve in the first quadrant.

Find the derivative of the function y = 3x - x^2 to determine the slope of the tangent line at any point (a, 3a - a^2). The derivative is dy/dx = 3 - 2x.

The equation of the tangent line at point (a, 3a - a^2) is y - (3a - a^2) = (3 - 2a)(x - a). Simplify this to find the y-intercept and x-intercept of the tangent line.

The x-intercept occurs when y = 0, and the y-intercept occurs when x = 0. Use these intercepts to determine the base and height of the triangle formed by the tangent line, the x-axis, and the y-axis.

Express the area of the triangle as a function of a, A(a) = (1/2) * base * height. Differentiate A(a) with respect to a, set the derivative equal to zero, and solve for a to find the value that minimizes the area.

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Key Concepts

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

Tangent Line to a Curve

A tangent line to a curve at a given point is a straight line that just touches the curve at that point. It has the same slope as the curve at the point of tangency. For the function y = 3x - x^2, the slope of the tangent line at any point (a, 3a - a^2) is given by the derivative, which is 3 - 2a.

Area of a Triangle

The area of a triangle can be calculated using the formula: Area = 0.5 * base * height. In this context, the triangle is formed by the x-axis, y-axis, and the tangent line. The base and height are determined by the x and y intercepts of the tangent line, which can be found using the point-slope form of the line equation.

Optimization in Calculus

Optimization involves finding the maximum or minimum value of a function. In this problem, we need to minimize the area of the triangle. This requires setting up an equation for the area in terms of a single variable, differentiating it, and finding critical points to determine the minimum area using the first or second derivative test.

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Related Practice

Textbook Question

Finding Critical Points

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

g(x) = √(2x − x²)

Textbook Question

Initial Value Problems

Solve the initial value problems in Exercises 71–90.

ds/dt = 1 + cos t, s(0) = 4

Textbook Question

Theory and Examples

In Exercises 53 and 54, show that the function has neither an absolute minimum nor an absolute maximum on its natural domain.

y = 3x + tan x

Textbook Question

22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.

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Textbook Question

Checking the Mean Value Theorem

Which of the functions in Exercises 7–12 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers.

f(x) = x⁴ᐟ⁵, [0, 1]

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 + tan²θ)dθ (Hint:1 + tan²θ = sec²θ)