Explanation Given : 3 x 2 + 6 x − 27 y − 51 = 0 2 x 2 + 2 y + 16 x + 8 = 0 The given equation, 3 x 2 + 6 x − 27 y − 51 = 0 First solve for y and then differentiate. 3 x 2 + 6 x − 27 y − 51 = 0 y = − 3 x 2 − 6 x + 51 − 27 Differentiate w .r .t x y ' = − 1 27 ( − 6 x − 6 ) For maximum or minimum, y ' = 0 − 1 27 ( − 6 x − 6 ) = 0 x = − 1 Now double differentiate of y y ' ' = 2 9 If y ' ' is negative at x = − 1 then maximum

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Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

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Chapter 4.6, Problem 2PS

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To find: The maximum value of x for which optimal value of y is obtained, 3x2+6x−27y−51=0

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Chapter 4.6, Problem 1PS

Chapter 4.6, Problem 3PS

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4 Consider f(x) periodic function with period 2, coinciding with (x) = -x on the interval [,0) and being the null function on the interval [0,7). The Fourier series of f: (A) does not converge in quadratic norm to f(x) on [−π,π] (B) is pointwise convergent to f(x) for every x = R П (C) is in the form - 4 ∞ +Σ ak cos(kx) + bk sin(kx), ak ‡0, bk ‡0 k=1 (D) is in the form ak cos(kx) + bk sin(kx), ak 0, bk 0 k=1

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Chapter 4 Solutions

Calculus

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The maximum value of x for which optimal value of y is obtained, 3 x 2 + 6 x − 27 y − 51 = 0

Solution Summary: The author explains that the given equation, 3x2+6x-27y-51=0 solves for y and then differentiates.

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To find: The maximum value of x for which optimal value of y is obtained, 3x2+6x−27y−51=0

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Chapter 4 Solutions