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The points you found are the vertices of the feasible region. I agree with the first three points you got. However, the last point should be (25/11, 35/11). This point is at the of the intersection of the two lines 8x-y = 15 and 3x+y = 10
So the four vertex points are:
(1,9)
(1,7)
(3,9)
(25/11, 35/11)
Plug each of those points, one at a time, into the objective function z = 7x+2y. The goal is to find the largest value of z
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Plug in (x,y) = (1,9)
z = 7x+2y
z = 7(1)+2(9)
z = 7+18
z = 25
We'll use this value later.
So let's call it A. Let A = 25
Plug in (x,y) = (1,7)
z = 7x+2y
z = 7(1)+2(7)
z = 7+14
z = 21
Call this value B = 21 so we can refer to it later
Plug in (x,y) = (3,9)
z = 7x+2y
z = 7(3)+2(9)
z = 21+18
z = 39
Let C = 39 so we can use it later
Finally, plug in (x,y) = (25/11, 35/11)
z = 7x+2y
z = 7(25/11)+2(35/11)
z = 175/11 + 70/11
z = 245/11
z = 22.2727 which is approximate
Let D = 22.2727
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In summary, we found
A = 25
B = 21
C = 39
D = 22.2727
The value C = 39 is the largest of the four results. This value corresponded to (x,y) = (3,9)
Therefore the max value of z is z = 39 and it happens when (x,y) = (3,9)
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Final Answer: 39
Answer:
Step-by-step explanation:
new length=15+x
width=19+x
then area=(15+x)×(19+x)=285+15x+19x+x²=x²+34x+285 ft²
original area=15×19=285 ft²
then 285+98=x²+34x+285
or
x²+34x-98=0
x²+34x+17²=98+17²
(x+17)²=98+289=387
x+17=√387=3√43
x=3√43-17 ft
What don't you get about the problem ?
