What is the minimum value for z =3x-1/2y over feasibility region defined by the constraints shown above
2 answers:
To obtain the minimum value of z, the first term of the binomial equation should be as small as possible, while the second term should be as large as possible. Thus, you must choose the lowest boundary of x which is 0. Then, choose the highest boundary of y which is 8. So, we substitute this to the equation:
So, the minimum value of z is -4.
So, the minimum value of z is -4.
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Let
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer
we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3
the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer
we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3
the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5