Answered: f(x,y) = x=y - 2xy? + 3xy +4 %3D

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter2: Functions And Graphs

Section2.6: Proportion And Variation

Problem 17E: Find the constant of proportionality. y is directly proportional to x. If x=30, then y=15.

See similar textbooks

Related questions

Concept explainers

Question

What is the global maximum and minimum of this function on the rectangle [0,2]x[0,2]?

Expert Solution

Step 1

Given function is $f (X, Y) = X^{2} Y - 2 X Y^{2} + 3 X Y + 4$

We have to find the global maximum and minimum of this function on the rectangle $[0, 2] x [0, 2]$

Differentiate $f (X, Y) = X^{2} Y - 2 X Y^{2} + 3 X Y + 4$ with respect to $X$

$\begin{matrix} f_{X} = \frac{d}{d X} (X^{2} Y - 2 X Y^{2} + 3 X Y + 4) \\ = 2 X Y - 2 Y^{2} + 3 Y + 0 \\ = 2 X Y - 2 Y^{2} + 3 Y \end{matrix}$

Hence, $f_{X} = 2 X Y - 2 Y^{2} + 3 Y$

Now, differentiate $f (X, Y) = X^{2} Y - 2 X Y^{2} + 3 X Y + 4$ with respect to $Y$

$\begin{matrix} f_{Y} = \frac{d}{d Y} (X^{2} Y - 2 X Y^{2} + 3 X Y + 4) \\ = X^{2} - 4 X Y + 3 X + 0 \\ = X^{2} - 4 X Y + 3 X \end{matrix}$

Hence, $f_{Y} = X^{2} - 4 X Y + 3 X$

Step 2

Equate $f_{X} = 0$ and $f_{Y} = 0$

$\begin{array}{l} \Rightarrow 2 X Y - 2 Y^{2} + 3 Y = 0..... (1) \\ \Rightarrow X^{2} - 4 X Y + 3 X = 0....... (2) \end{array}$

Now, solve equation (1)

$\begin{array}{l} \Rightarrow 2 X Y - 2 Y^{2} + 3 Y = 0 \\ \Rightarrow Y (2 X - 2 Y + 3) = 0 \\ \Rightarrow Y = 0 or 2 X - 2 Y + 3 = 0 \\ \Rightarrow Y = 0 or Y = \frac{2 X + 3}{2} \end{array}$

Now, solve equation (2)

$\begin{array}{l} \Rightarrow X^{2} - 4 X Y + 3 X = 0 \\ \Rightarrow X (X - 4 Y + 3) = 0 \\ \Rightarrow X = 0 or X - 4 Y + 3 = 0 \\ \Rightarrow X = 0 or X = 4 Y - 3 \end{array}$

Now, substitute $X = 4 Y - 3$ in $Y = \frac{2 X + 3}{2}$

$\begin{array}{l} \Rightarrow Y = \frac{2 (4 Y - 3) + 3}{2} \\ \Rightarrow Y = \frac{8 Y - 6 + 3}{2} \\ \Rightarrow Y = \frac{8 Y - 3}{2} \\ \Rightarrow 2 Y = 8 Y - 3 \\ \Rightarrow Y = \frac{1}{2} \end{array}$

Now, substitute $Y = \frac{1}{2}$ in $X = 4 Y - 3$

$\begin{array}{l} \Rightarrow X = 4 (\frac{1}{2}) - 3 \\ \Rightarrow X = - 1 \end{array}$

Hence, critical points are $(0, 0)$ , $(0, \frac{1}{2})$ , $(- 1, 0)$ and $(- 1, \frac{1}{2})$

Since, we have to choose global maximum and minimum in $[0, 2] x [0, 2]$

Hence, critical points are $(0, 0)$ and $(0, \frac{1}{2})$ .

Step by step

Solved in 3 steps

Knowledge Booster

$Application of Differentiation$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.