Who did I get the boundaries wrong? Find the exact extreme values of the function: = f (x, y) = 3x2 – 2.ry + 4y? – 6x – 20y +5subject to the following constraints:0

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Description: Who did I get the boundaries wrong? Find the exact extreme values of the function: = f (x, y) = 3x2 – 2.ry + 4y? – 6x – 20y +5subject to the following constraints:0

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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Who did I get the boundaries wrong?

Find the exact extreme values of the function
: = f (x, y) = 3x2 – 2.ry + 4y? – 6x – 20y +5
subject to the following constraints:
0 <r < 5
0 < y<6
Start by listing all nine candidates, including their z values, in the form (x,y,z):
First, list the four corner points and order your answers from smallest to largest x, then from smallest to largest y.
1) ( 0
5
2) ( 0
29
3) ( 5
50
4) ( 5
6
14
Next find the critical point.
5) ( 2
3
-31
Lastly, find the four boundary points and order your answers from smallest to largest x, then from smallest to largest y.
6) ( 0
5/2
25/2
7) ( 5
15/4
-25/4 X )
8) ( 1
2
9) ( 3
2
X X
CO
X X

Expert Solution

Step 1

Given function is $z = 3 x^{2} - 2 x y + 4 y^{2} - 6 x - 20 y + 5$

We have to find the four boundary points and order your answer from smallest to largest $x$ .

(6) On boundary $x = 0$

$\begin{matrix} z = 3 {(0)}^{2} - 2 (0) y + 4 y^{2} - 6 (0) - 20 y + 5 \\ = 4 y^{2} - 20 y + 5 \end{matrix}$

Hence, $z = 4 y^{2} - 20 y + 5$

Differentiate $z = 4 y^{2} - 20 y + 5$ with respect to $y$

$\begin{matrix} z_{y} = \frac{d}{d y} (4 y^{2} - 20 y + 5) \\ = 8 y - 20 \end{matrix}$

Therefore, $z_{y} = 8 y - 20$

Equate $z_{y} = 0$

$\begin{array}{l} \Rightarrow 8 y - 20 = 0 \\ \Rightarrow 8 y = 20 \\ \Rightarrow y = \frac{20}{8} \\ \Rightarrow y = \frac{5}{2} \end{array}$

Hence, boundary points is $x = 0$ and $y = \frac{5}{2}$

Substitute $x = 0$ and $y = \frac{5}{2}$ in $z = 3 x^{2} - 2 x y + 4 y^{2} - 6 x - 20 y + 5$

$\begin{matrix} z = 3 {(0)}^{2} - 2 (0) (\frac{5}{2}) + 4 {(\frac{5}{2})}^{2} - 6 (0) - 20 (\frac{5}{2}) + 5 \\ = 0 - 0 + 25 - 0 - 50 + 5 \\ = - 20 \end{matrix}$

Hence, $z = - 20$

Therefore, point $(x, y, z) = (0, \frac{5}{2}, - 20)$

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