Find The Maximum/Minimum Value Y=-x^2+4x-4

Find the maximum and minimum values of f (x, y) = x 3 + 2 xy + 4 y 2 subject to x + 2 y = 12. Be sure to use the method at the end of Example 1 to determine whether each solution is a maximum or a minimum.The line 5x +2y = 200 meets the coordinate axes at C(40,0) and D(0, 100) respectively. By joining these points we obtain the line 5x +2y = 200.Clearly (0,0) does not satisfies the inequation 5x +2y ≥ 200. So,the region which does not contain the origin represents the solution set of the inequation 5x +2y ≥ 200. Region represented by x +2y(3 points) Constraints {x>0 {y>0 {-x+3>y {y . Algebra ll . By graphing a system of constraints, find the values of x and y that maximize the objective function. x+y less than it equal to 8 2x + y is less than it equal to 10 x is great her than it equal to 0 y is great her than it equal to . MathematicsWhat is the maximum value of P = 4x + 2y, given the constraints on x and y listed below? 2 See answers The maximum value of P is 40. Where x is 10 and y is 0. fiddykal32 fiddykal32 Answer: see pic below. Which expression is equivalent to ? Assume .Since the graph of P = 2x + y is linear (planal), the point (x, y) at which the maximum value of P occurs will be found along at a corner of the area defined by the constraints. If we graph the constraints, we find that there are two corners at the intersections of 3x + 14y = 1568 & 8x + 10y = 1448 and 8x + 10y = 1448 & 17x + 3y = 1836.

The Maximum Value of Z = 4x + 3y Subjected to the

What is the maximum value of P = 4x + 2y, given the constraints on x and y listed below? D. 40. The constraints of a problem are listed below. What are the vertices of the feasible region? 2x + 3y > 12 5x + 2y > 15 x > 0 y >0. Store Y can provide a maximum of 40 bags per week, and the farm has committed to buy at least as many bags fromWhat is the maximum value of P = 4x + 2y, given the constraints on x and y listed below? x + 2y < or equal to 10 y < or equal to 2 x > or equal to 0 y > or equal to 0. 40. The constraints of a problem are listed below. What are the vertices of the feasible region? 2x + 3y > or equal to 12Clearly 6= 0. It follows that x= 3 and y= 4 . Using the constraint, x 2+ y = 25 9 2 + 16 = 25 25 2 = 25 2 = 1 = 1: The points of interest are (x;y) = (3;4). Therefore, the maximum and mimimum valuesWhat is the maximum value of P = 4x + 2y, given the constraints on x and y listed below? x + 2y < or equal to 10 y < or equal to 2 x > or equal to 0 y > or equal to 0. 40. The constraints of a problem are listed below. What are the vertices of the feasible region? 2x + 3y > or equal to 12

What point maximizes the objective function P = 4x + 3y

The maximum (if it exists) is the largest value of P at a vertex. The minimum is the smallest value of P at a vertex. If the objective function is maximized (or minimized) at two vertices, it is minimized (or maximized) at every point connecting the two vertices. Example: Find the maximum and minimum values of P=3x+2y subject to x + 4y ≤ 20What is the maximum value of P = 24x + 30y, given the constraints on x and y listed below? x+y (less than or equal to) 5 x-y (greater than or equal to) -1 x (greater than or equal to) 0 y (greater than or equal to) 0 120 132 138 150The maximum value of the object function Z = 5x + 10 y subject to the constraints x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0 is (a) 300 (b) 600 (c) 400 (d) 800. Answer. Answer: (b) 600The maximum value of P will be found where x=10, y=0. That maximum is.. Pmax = 4*10 +2*0 = 40 _____ The objective function puts the greater value on x, so it will be maximized at the maximum possible value of x.The constraints of a problem are listed below. What are the vertices of the feasible region? x+y >= 7 x-2y <= -2 x >= 0 y >= 0

P < 28

Step-by-step explanation:

given the constraints:

x + 2y < 10

y < 2

x>0

y>0

we can see that the vary of y is from 0 to 2:

we will use the maximum value of y in x+2y<10, to find the vary of x

x + 2y < 10

x + 2(2) < 10

x < 10 -4

x < 6

now we know that vary of x is from 0 to six:

take a more in-depth have a look at those ranges, each of the levels of x and y do not include their excessive values. (these don't seem to be )

The serve as P is:

P = 4x + 2y, we can simply installed all the maximum ranges of x and y in the equation to seek out the maximum value of P. however earlier than plugging in the values we will have to watch out: we're putting the extremes of the ranges in the equation, but these extremes aren't in the precise vary itself.

so as a substitute of writing that the maximum value of P is equivalent to a host, we must write that the maximum value of P is on the subject of that number(or approaching that number from below)

P < 4(6) + 2(2)

P < 28

NOTE:

if the constraints given include the extremes of the ranges. Then the resolution will probably be P = 28.

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