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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<h3>Answer:</h3>
- f(x) = -2x^3 +3x^2 +11x -6
- see attached
- an infinite number. Since the magnitude of the leading coefficient is not specified, it may be any negative number. (We have chosen the smallest magnitude integer that makes all coefficients be integers.)
1. When "a" is a root of a polynomial, (x -a) is a factor of it. For the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).
In order to make the leading coefficient be negative, we need to multiply this product by a negative number. Any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.
Then ...
... f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)
... = -(2x² +3x -2)(x -3)
... = -(2x³ -3x² -11x +6)
... f(x) = -2x³ +3x² +11x -6
2. A graph created by the Desmos on-line graphing calculator is shown, and the zeros are highlighted.
3. As indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. You want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.