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# Math help?

Maximize the function

f(x, y) = 6x + 7y

in the region determined by the following constraints.

3x+ 2y ≤ 18

3x + 4y≥ 12

x≥0

y≥0

### 3 Answers

- PinkgreenLv 76 months agoFavorite Answer
Maximize f=6x+7y

constraints:

3x+2y<=18

3x+4y>=12

x, y>=0

(a) Graphical solutions:

(i) draw lines

3x+2y=18----(1)

3x+4y=12----(2)

7y+6x=0=>y=-6x/7----(3)

on a piece of the graph paper.

(ii) Find out the feasible region R

which is bounded by (1), (2) &

x=0, y=0.

(iii) Push the line (3) parallely to the

farthest corner of R which should be

(0,9) with the help of the set squares.

(iv) The max.f=6(0)+7(9)=63.

(b) The numerical method:

(i) Using the M-technique with M=1000.

Rewrite the constraints as

3x+2y+S1=18

3x+4y-S2+R=12

x, y>=0

Maximize f-3006x-4007y+1000S2=-12000.

After 2 iterations, get

x=0

y=9

S1=0

S2=24

R=0

Max. f=63

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- Φ² = Φ+1Lv 76 months ago
Please check answers already provided here: https://au.answers.yahoo.com/question/index?qid=20...

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- husoskiLv 76 months ago
It's a linear function, so it will be maximized on the boundary; specifically at one of the vertices when the bounary is a polygon. Write the first two inequalities as equations in intercept form:

x/6 + y/9 = 1 . . . . divide by 18 to get x-intercept = 6 and y-intercept = 9

x/4 + y/3 = 1 . . . . divide by 12 to get x-intercept = 4, y intercept = 3

The >= direction of the 2nd inequality, and the fact both intercepts of the second boundary line are inside the region defined by the 1st inequality, make it easy. All corners of the boundary are on the axes. Evaluate the function at (6, 0), (0, 9) , (4, 0) and (0, 4) to find the maximum location and value. I get f(0, 9) = 63.

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