To find the minimum value of (2x + 2y) in a feasible region, you typically need to know the constraints that define that region. If you have a specific set of inequalities or constraints, you can apply methods like the corner point theorem or linear programming techniques to evaluate the objective function at the vertices of the feasible region. Without specific constraints, it's impossible to determine the minimum value accurately. If you provide the constraints, I can assist you further in finding the minimum.
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.
To find the maximum value of (2x + 2y) in the feasible region, you typically need to identify the constraints that define this region, often in the form of inequalities. Then, you would evaluate the objective function at the vertices of the feasible region, which are the points of intersection of the constraints. The maximum value will be found at one of these vertices. If you provide the specific constraints, I can help you calculate the maximum value.
To maximize ( x^3y^4 ) given the constraint ( 2x + 3y = 7 ) and ( x \geq 0, y \geq 0 ), we can use the method of Lagrange multipliers or substitute ( y ) in terms of ( x ). From the equation, express ( y ) as ( y = \frac{7 - 2x}{3} ). Substituting this into ( x^3y^4 ) will yield a function of ( x ) that can be maximized within the feasible region defined by the constraints. Solving this will give the maximum value of ( x^3y^4 ).
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.
2x+2y
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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.
To find the maximum value of (2x + 2y) in the feasible region, you typically need to identify the constraints that define this region, often in the form of inequalities. Then, you would evaluate the objective function at the vertices of the feasible region, which are the points of intersection of the constraints. The maximum value will be found at one of these vertices. If you provide the specific constraints, I can help you calculate the maximum value.
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To maximize ( x^3y^4 ) given the constraint ( 2x + 3y = 7 ) and ( x \geq 0, y \geq 0 ), we can use the method of Lagrange multipliers or substitute ( y ) in terms of ( x ). From the equation, express ( y ) as ( y = \frac{7 - 2x}{3} ). Substituting this into ( x^3y^4 ) will yield a function of ( x ) that can be maximized within the feasible region defined by the constraints. Solving this will give the maximum value of ( x^3y^4 ).
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)
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(2x)2 = 4 x2 Its numerical value depends on the value of 'x'.