Find the maximum and the minimum value of the following objective function, and the value of x and y at which they occur. The f
1 answer:
Answer:
The maximum value of the objective function is 112 when x = 0 and y = 7.
Step-by-step explanation:
Given the constraints:
5x+3y≤37, 3x+5y≤35, x≥0, y≥0
Plotting the above constraints using geogebra online graphing tool, we get the solution to the constraints as:
A(0, 7), B(7.4, 0), C(5, 4) and D(0, 0)
The objective function is given as E =2x+16y, therefore:
At point A(0, 7): E = 2(0) + 16(7) = 112
At point B(7.4, 0): E = 2(7.4) + 16(0) = 14.8
At point C(5, 4): E = 2(5) + 16(4) = 74
At point D(0, 0): E = 2(0) + 16(0) = 0
Therefore the maximum value of the objective function is at A(0, 7).
The maximum value of the objective function is 112 when x = 0 and y = 7.
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0=ax²+bx+c
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mins 9 from both sides and minus x
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a=7
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so the 4th one
so
for
0=ax²+bx+c
so
7x²=9+x
mins 9 from both sides and minus x
7x²-1x-9=0
a=7
b=-1
c=-9
so the 4th one
Plot these points
vertex: (-4,-7)
x intercepts: (-5.5,0), (-2.5, 0)
and those two other points
the graph should look something like this:
vertex: (-4,-7)
x intercepts: (-5.5,0), (-2.5, 0)
and those two other points
the graph should look something like this: