The binding constraints for this problem are the first and second. Min x1 + 2x2 s. t. x1 + x2 ≥ 300 2x1 + x2 ≥ 400 2x1 + 5x2 ≤ 750 x1, x2 ≥ 0 a. Keeping c2 fixed at 2, over what range can c1 vary before there is a change in the optimal solution point? b. Keeping c1 fixed at 1, over what range can c2 vary before there is a change in the optimal solution point? c. If the objective function becomes Min 1.5x1 + 2x2, what will be the optimal values of x1, x2, and the objective function? d. If the objective function becomes Min 7x1 + 6x2, what constraints will be binding? e. Find the dual price for each constraint in the original problem.

d? maybe

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180

hello i might . what do you need.

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let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0