Find the maximum and the minimum value of the following objective​ function, and the value of x and y at which they occur. The function F=2x+16y subject to 5x+3y≤37, 3x+5y≤35, x≥0, y≥0 The maximum value of the objective function is ___ when x=___ and y=___

The maximum value of the objective function is 112 when x = 0 and y = 7.

Step-by-step explanation:

Given the constraints:

5x+3y≤37, 3x+5y≤35, x≥0, y≥0

Plotting the above constraints using geogebra online graphing tool, we get the solution to the constraints as:

A(0, 7), B(7.4, 0), C(5, 4) and D(0, 0)

The objective function is given as E =2x+16y, therefore:

At point A(0, 7):  E = 2(0) + 16(7) = 112

At point B(7.4, 0): E = 2(7.4) + 16(0) = 14.8

At point C(5, 4): E = 2(5) + 16(4) = 74

At point D(0, 0): E = 2(0) + 16(0) = 0

Therefore the maximum value of the objective function is at A(0, 7).

The maximum value of the objective function is 112 when x = 0 and y = 7.

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c

1260=3*6*7*10=3^2*2^2*7*5 perfect cude = 3^3*2^

step-by-step explanation: