Solved: Find the absolute maximum and minimum values of

Chapter 13, Problem 62E

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Extreme Values on Parametrized Curves To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t)\), we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the

a. critical points (points where \(d f / d t\) is zero or fails to exist), and

b. endpoints of the parameter domain.

Find the absolute maximum and minimum values of the following functions on the given curves.

Functions:

a. \(f(x, y)=2 x+3 y\)          b. \(g(x, y)=x y\)

c. \(h(x, y)=x^{2}+3 y^{2}\)

Curves:

i) The semiellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1, \quad y \geq 0\)

ii) The quarter ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1, \quad x \geq 0, \quad y \geq 0\)

Use the parametric equations \(x=3 \cos t, y=2 \sin t\)

Equation Transcription:

f(x,y)

x=x(t), y=y(t)

f

t

df/dt

f(x,y)=2x+3y

g(x,y)=xy

h(x,y)=x2+3y2

(x2/9)+(y2/4)=1, y≥0

(x2/9)+(y2/4)=1, x≥0, y≥0

x=3 cos⁡ t, y=2 sin⁡ t

Text Transcription:

f(x,y)

x=x(t), y=y(t)

f

t

df/dt

f(x,y)=2x+3y

g(x,y)=xy

h(x,y)=x^2+3y^2

(x^2/9)+(y^2/4)=1, y≥0

(x^2/9)+(y^2/4)=1, x≥0, y≥0

x=3 cos⁡ t, y=2 sin⁡ t