Explanation Given information: The product of two positive numbers is 784. Calculation: The product of two positive numbers is 784. (Given) ⇒ x y = 784 Its sum is denoted by S ( x ) . ⇒ S ( x ) = x + y S ( x ) = x + 784 x Differentiating with respect to ‘x’, we get S ' ( x ) = 1 − 784 x 2 To obtain the critical point, put S ' ( x ) = 0 . ⇒ 1 − 784 x 2 = 0 ⇒ 784 x 2 = 1 ⇒ x 2 = 784 ⇒ x = ± 28 Since x is a positive value, we ignore the negative sign Thus, x = 28

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Chapter 4.3, Problem 36P

To determine

To find:

The minimum value of the sum of two numbers.

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Chapter 4.3, Problem 35P

Chapter 4.3, Problem 37P

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Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

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Chapter 4 Solutions

Applied Calculus, 6e WileyPLUS + Loose-leaf

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The minimum value of the sum of two numbers.

Solution Summary: The author explains that the minimum value of the sum of two numbers is 56. The sum is denoted by S(x).

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To find: The minimum value of the sum of two numbers.

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Chapter 4 Solutions

To find:

The minimum value of the sum of two numbers.