To find the maximum value of 2x + 5y within the feasible region, you would need to evaluate the objective function at each corner point of the feasible region. The corner points are the vertices of the feasible region where the constraints intersect. Calculate the value of 2x + 5y at each corner point and identify the point where it is maximized. This point will give you the maximum value of 2x + 5y within the feasible region.
That completely depends on the values of 'x' and 'y'. The value of that expression at any instant is exactly double the sum of 'x' and 'y', but as soon as either of them changes, the value of the expression likewise instantly changes. The only possible statement is that the value of the expression is minimum when the sum of 'x' and 'y' is minimum.
No. For 0 < x < 2, 2x is larger.
If you mean 2x = 28 then the value of x is 14
y = x2 - 2x + 10 This is a minimum when dy/dx = 0 dy/dx = 2x - 2 : When 2x - 2 = 0 then 2x = 2 : x = 1 Substituting in the equation gives y = 12 - (2 x 1) + 10 = 1 - 2 + 10 = 9.
2x+2y
14
To find the maximum value of 2x + 5y within the feasible region, you would need to evaluate the objective function at each corner point of the feasible region. The corner points are the vertices of the feasible region where the constraints intersect. Calculate the value of 2x + 5y at each corner point and identify the point where it is maximized. This point will give you the maximum value of 2x + 5y within the feasible region.
The answer obviously depends on what the boundaries of the feasibility region are.
5
That completely depends on the values of 'x' and 'y'. The value of that expression at any instant is exactly double the sum of 'x' and 'y', but as soon as either of them changes, the value of the expression likewise instantly changes. The only possible statement is that the value of the expression is minimum when the sum of 'x' and 'y' is minimum.
dy/dx= 2x therefore d2ydx2= 2 as this is positive we can tell that it is the minimum value in a curve
The minimum value of the parabola is at the point (-1/3, -4/3)
(2x)2 = 4 x2 Its numerical value depends on the value of 'x'.
18
No. For 0 < x < 2, 2x is larger.
If you mean 2x = 28 then the value of x is 14