Solve the following linear program using the graphical solution procedure: Max 5A + 5B s.t. 1A lessthanorequalto 100 1B lessthan
orequalto 80 2A + 4B lessthanorequalto 400 A, B greaterthanorequalto 0
1 answer:
<h2>Answer:</h2>
- The optimal solution to the given linear programming problem exist at (100,50)
- and the optimal solution is: 5A+5B= 750
We are given a system of linear programming problem as follows:
Max 5A + 5B
s.t. A ≤ 100---------------(1)
B ≤ 80-------------(2)
2A+4B ≤400-------------(3)
which is given by:
A+2B ≤ 200
and A,B≥0
This means that the solution to this LPP will lie in the first quadrant.
( Since, both the variables A and B are greater than or equal to zero)
Now, we consider that A is represented by the x axis and B by y-axis.
We know that the optimal solution always exist at the boundary point.
Hence, by plotting these inequalities in the graph we get the boundary points as:
(0,0) , (0,80) , (100,0) , (100,50) and (40,80)
Now, we will check at which boundary point the optimal function is maximized .
Point Value of optimal function( 5A+5B)
(0,0) 0
(0,80) 400
(100,0) 500
(100,50) 750
(40,80) 600
The maximum value is obtained at (100,50).
and the value of the optimal solution is: 750
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Answer:
The coordinates of image are A'(-2,1), B'(1,0) and C'(-1,0).
Step-by-step explanation:
From the figure it is clear that the coordinates of triangle are A(0,0), B(1,3) and C(1,1).
∆ABC is translated 2 units down and 1 unit to the left.
Then it is rotated 90° clockwise about the origin to form ∆A′B′C′.
Therefore the coordinates of image are A'(-2,1), B'(1,0) and C'(-1,0).
Answer:
34°
Step-by-step explanation:
If m∠ADE is with 34° smaller than m∠CAB, then denote
m∠ADE=x°,
m∠CAB=(x+34)°.
Since DE ║ AB, then
m∠ADE=m∠DAB=x°.
AD is angle A bisector, then
m∠EAD=m∠DAB=x°.
Thus,
m∠CAB=m∠CAD+m∠DAB=(x+x)°=2x°.
On the other hand,
m∠CAB=(x+34)°,
then
2x°=(x+34)°,
m∠ADE=x°=34°.
Answer:
a) P(X<50)=0.9827
b) P(X>47)=0.4321
c) P(-1.5<z<1.5)=0.8664
Step-by-step explanation:
We will calculate the probability based on a random sample of one moped out of the population, normally distributed with mean 46.7 and standard deviation 1.75.
a) This means we have to calculate P(x<50).
We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:
b) We have to calculatee P(x>47).
We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:
c) If the value differs 1.5 standard deviations from the mean value, we have a z-score of z=1.5
So the probability that maximum speed differs from the mean value by at most 1.5 standard deviations is P(-1.5<z<1.5):