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The question cannot be answered because:

  • there is no symbol shown between 2y and x,
  • there is no information on the feasible region.



The question cannot be answered because:

  • there is no symbol shown between 2y and x,
  • there is no information on the feasible region.



The question cannot be answered because:

  • there is no symbol shown between 2y and x,
  • there is no information on the feasible region.



The question cannot be answered because:

  • there is no symbol shown between 2y and x,
  • there is no information on the feasible region.
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11y ago
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11y ago

The question cannot be answered because:

  • there is no symbol shown between 2y and x,
  • there is no information on the feasible region.
This answer is:
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Q: What POINT identifies the maximum value of 2y x in the feasible region?
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Why the derivative is set equal to zero?

The first derivative is set to zero to find the critical points of the function. A critical point can be a minimum, maximum, or a saddle point. There's a reason for this. Suppose a differentiable function f:R->R has a maximum at x=a. Then the function goes down to the right of a, which means f'(a)


What is the relationship between the Green's Theorem Divergence Theorem and Stoke's Theorem?

GREEN'S THEOREM: if m=m(x,y) and n= n(x,y) are the continuous functions and also partial differential in a region 'r' of x,y plane bounded by a simple closed curve c. DIVERGENCE THEOREM: if f is a vector point function having continuous first order partial derivatives in the region v bounded by a closed curve s


How do you graph x is greater than or equal to y?

Take a blank graph with 'x' and 'y' axes on it. Draw a 45-degree line on the graph. The line goes through the origin, and from the origin, it goes down-left and up-right. The slope of the line is 1, and its equation is y=x. The region "y is greater than or equal to x" is every point on that line, plus every point on the side above it (to the left of it).


Identify the maximum value of the function y equals -6x2-12x-1?

y = - 6x2 - 12x - 1 A second degree equation graphs as a parabola, and has only one max or min. At that point, the first derivative y' = 0. dy/dx = - 12x - 12 = 0 - x - 1 = 0 ==> x = - 1 At that point, y = - 6( 1 ) - 12( - 1 ) - 1 = - 6 + 12 - 1 = 5. The max value of the function is 5, and occurs when x = -1.


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Related questions

What is the value of 6x 5y at point D in the feasible region?

Given definitions, or descriptions at least, of "point D" and "the feasible region",I might have had a shot at answering this one.


What is the minimum value of 6x plus 5y in the feasible region excluding point (0 0)?

The answer depends on the feasible region and there is no information on which to determine that.


What is the maximum value of 3x + 3y in the feasible region?

To find the maximum value of 3x + 3y in the feasible region, you will need to determine the constraints on the variables x and y and then use those constraints to define the feasible region. You can then use linear programming techniques to find the maximum value of 3x + 3y within that feasible region. One common way to solve this problem is to use the simplex algorithm, which involves constructing a tableau and iteratively improving a feasible solution until an optimal solution is found. Alternatively, you can use graphical methods to find the maximum value of 3x + 3y by graphing the feasible region and the objective function 3x + 3y and finding the point where the objective function is maximized. It is also possible to use other optimization techniques, such as the gradient descent algorithm, to find the maximum value of 3x + 3y within the feasible region. Without more information about the constraints on x and y and the specific optimization technique you wish to use, it is not possible to provide a more specific solution to this problem.


Why do some feasible regions have more corners than other feasible regions?

Feasible regions have more corners when there are more constraints that intersect at a single point, creating a corner. If there are more constraints that intersect at different points, the feasible region will have more corners. In general, the number of corners in a feasible region is determined by the number of constraints and how they interact.


What point in the feasible region maximize C for the objective function C 4x - 2y?

1y


What point in the feasible region maximizes C for the objective function C equals 4x-2y?

1y


Why optimal solution is only at corner point?

feasible region gives a solution but not necessarily optimal . All the values more/better than optimal will lie beyond the feasible .So, there is a good chance that the optimal value will be on a corner point


What is optimal solution?

It is usually the answer in linear programming. The objective of linear programming is to find the optimum solution (maximum or minimum) of an objective function under a number of linear constraints. The constraints should generate a feasible region: a region in which all the constraints are satisfied. The optimal feasible solution is a solution that lies in this region and also optimises the obective function.


What is the feasible region in linear programming?

Linear programming is just graphing a bunch of linear inequalities. Remember that when you graph inequalities, you need to shade the "good" region - pick a point that is not on the line, put it in the inequality, and the it the point makes the inequality true (like 0


Which point gives the minimum value p3x-2y in the feasible region shown?

If you want to ask questions about the "region shown", then it would have helped if you could make sure that there is some region that is shown. However, given the limitations of the browser used by this site, you do not have much of a hope!


What decimal identifies point D?

The answer depends on where or what point D is.


How do you know whether to use an open circle or a closed circle when graphing an inequality and why?

An open or closed circle are used to graph an inequality in one variable. An open circle is used if the value at the end point is excluded from the feasible region while a closed circle is used if the value at that point is within the accepted region.