Question 1187741
<br>
Nearly all resources on linear programming will say that you need to evaluate the objective function at all corners of the feasibility region to find the minimum and maximum values.<br>
That is not true.<br>
You can find the corners of the feasibility region that give the minimum and maximum values of the objective function by looking at the slope of the objective function.<br>
Here is the nice drawing of the feasibility region that I pirated from Edwin's response:<br>
{{{drawing(400,8800/27,-.9,9.9,-.9,7.9, 
graph(400,8800/27,-.9,9.9,-.9,7.9), line(-9,14,18,-7),
line(-3,4,12,4),line(7,-12,7,12),line(0,-3,0,12),line(-3,0,12,0),
green(
line(0,4,0,0),line(0.07,4,0.07,0),line(0.14,4,0.14,0),line(0.14,4,0.14,0),
line(0.21,4,0.21,0),line(0.28,4,0.28,0),line(0.35,4,0.35,0),line(0.35,4,0.35,0),
line(0.42,4,0.42,0),line(0.49,4,0.49,0),line(0.56,4,0.56,0),line(0.56,4,0.56,0),
line(0.63,4,0.63,0),line(0.7,4,0.7,0),line(0.77,4,0.77,0),line(0.77,4,0.77,0),
line(0.84,4,0.84,0),line(0.91,4,0.91,0),line(0.98,4,0.98,0),line(0.98,4,0.98,0),
line(1.05,4,1.05,0),line(1.12,4,1.12,0),line(1.19,4,1.19,0),line(1.19,4,1.19,0),
line(1.26,4,1.26,0),line(1.33,4,1.33,0),line(1.4,4,1.4,0),line(1.4,4,1.4,0),
line(1.47,4,1.47,0),line(1.54,4,1.54,0),line(1.61,4,1.61,0),line(1.61,4,1.61,0),
line(1.68,4,1.68,0),line(1.75,4,1.75,0),line(1.82,4,1.82,0),line(1.82,4,1.82,0),
line(1.89,4,1.89,0),line(1.96,4,1.96,0),line(2.03,4,2.03,0),line(2.03,4,2.03,0),
line(2.1,4,2.1,0),line(2.17,4,2.17,0),line(2.24,4,2.24,0),line(2.24,4,2.24,0),
line(2.31,4,2.31,0),line(2.38,4,2.38,0),line(2.45,4,2.45,0),line(2.45,4,2.45,0),
line(2.52,4,2.52,0),line(2.59,4,2.59,0),line(2.66,4,2.66,0),line(2.66,4,2.66,0),
line(2.73,4,2.73,0),line(2.8,4,2.8,0),line(2.87,4,2.87,0),line(2.87,4,2.87,0),
line(2.94,4,2.94,0),line(3.01,4,3.01,0),line(3.08,4,3.08,0),line(3.08,4,3.08,0),
line(3.15,4,3.15,0),line(3.22,4,3.22,0),line(3.29,4,3.29,0),line(3.29,4,3.29,0),
line(3.36,4,3.36,0),line(3.43,4,3.43,0),line(3.5,4,3.5,0),line(3.5,4,3.5,0),
line(3.57,4,3.57,0),line(3.64,4,3.64,0),line(3.71,4,3.71,0),line(3.71,4,3.71,0),
line(3.78,4,3.78,0),line(3.85,4,3.85,0),

line(3.86,0,3.86,3.99777778),line(3.93,0,3.93,3.94333333),line(4.0,0,4.0,3.88888889),line(4.0,0,4.0,3.88888889),
line(4.07,0,4.07,3.83444444),line(4.14,0,4.14,3.78),line(4.21,0,4.21,3.72555556),line(4.21,0,4.21,3.72555556),
line(4.28,0,4.28,3.67111111),line(4.35,0,4.35,3.61666667),line(4.42,0,4.42,3.56222222),line(4.42,0,4.42,3.56222222),
line(4.49,0,4.49,3.50777778),line(4.56,0,4.56,3.45333333),line(4.63,0,4.63,3.39888889),line(4.63,0,4.63,3.39888889),
line(4.7,0,4.7,3.34444444),line(4.77,0,4.77,3.29),line(4.84,0,4.84,3.23555556),line(4.84,0,4.84,3.23555556),
line(4.91,0,4.91,3.18111111),line(4.98,0,4.98,3.12666667),line(5.05,0,5.05,3.07222222),line(5.05,0,5.05,3.07222222),
line(5.12,0,5.12,3.01777778),line(5.19,0,5.19,2.96333333),line(5.26,0,5.26,2.90888889),line(5.26,0,5.26,2.90888889),
line(5.33,0,5.33,2.85444444),line(5.4,0,5.4,2.8),line(5.47,0,5.47,2.74555556),line(5.47,0,5.47,2.74555556),
line(5.54,0,5.54,2.69111111),line(5.61,0,5.61,2.63666667),line(5.68,0,5.68,2.58222222),line(5.68,0,5.68,2.58222222),
line(5.75,0,5.75,2.52777778),line(5.82,0,5.82,2.47333333),line(5.89,0,5.89,2.41888889),line(5.89,0,5.89,2.41888889),
line(5.96,0,5.96,2.36444444),line(6.03,0,6.03,2.31),line(6.1,0,6.1,2.25555556),line(6.1,0,6.1,2.25555556),
line(6.17,0,6.17,2.20111111),line(6.24,0,6.24,2.14666667),line(6.31,0,6.31,2.09222222),line(6.31,0,6.31,2.09222222),
line(6.38,0,6.38,2.03777778),line(6.45,0,6.45,1.98333333),line(6.52,0,6.52,1.92888889),line(6.52,0,6.52,1.92888889),
line(6.59,0,6.59,1.87444444),line(6.66,0,6.66,1.82),line(6.73,0,6.73,1.76555556),line(6.73,0,6.73,1.76555556),
line(6.8,0,6.8,1.71111111),line(6.87,0,6.87,1.65666667),line(6.94,0,6.94,1.60222222),line(6.94,0,6.94,1.60222222)), line(0,-3,0,12),line(-3,0,12,0)

 )}}}<br>
The objective function in this problem (which, by the way, is not a reasonable one) is<br>
P = 17x - 4y​ + 61<br>
In slope-intercept form, that equation is<br>
y = (17/4)x + (61-P)/4<br>
So the slope of the objective function is 17/4.<br>
<hr>
***** THE MINIMUM AND MAXIMUM VALUES OF THE OBJECTIVE FUNCTION WILL BE OBTAINED WHERE LINES WITH SLOPES OF 17/4 JUST TOUCH THE FEASIBILITY REGION. *****<br>
<hr>
Here is the graph of the feasibility region again, with a few lines with slope 17/4 added:<br>
{{{drawing(400,8800/27,-.9,9.9,-.9,7.9, 
graph(400,8800/27,-.9,9.9,-.9,7.9), line(-9,14,18,-7),
line(-3,4,12,4),line(7,-12,7,12),line(0,-3,0,12),line(-3,0,12,0),
green(
line(0,4,0,0),line(0.07,4,0.07,0),line(0.14,4,0.14,0),line(0.14,4,0.14,0),
line(0.21,4,0.21,0),line(0.28,4,0.28,0),line(0.35,4,0.35,0),line(0.35,4,0.35,0),
line(0.42,4,0.42,0),line(0.49,4,0.49,0),line(0.56,4,0.56,0),line(0.56,4,0.56,0),
line(0.63,4,0.63,0),line(0.7,4,0.7,0),line(0.77,4,0.77,0),line(0.77,4,0.77,0),
line(0.84,4,0.84,0),line(0.91,4,0.91,0),line(0.98,4,0.98,0),line(0.98,4,0.98,0),
line(1.05,4,1.05,0),line(1.12,4,1.12,0),line(1.19,4,1.19,0),line(1.19,4,1.19,0),
line(1.26,4,1.26,0),line(1.33,4,1.33,0),line(1.4,4,1.4,0),line(1.4,4,1.4,0),
line(1.47,4,1.47,0),line(1.54,4,1.54,0),line(1.61,4,1.61,0),line(1.61,4,1.61,0),
line(1.68,4,1.68,0),line(1.75,4,1.75,0),line(1.82,4,1.82,0),line(1.82,4,1.82,0),
line(1.89,4,1.89,0),line(1.96,4,1.96,0),line(2.03,4,2.03,0),line(2.03,4,2.03,0),
line(2.1,4,2.1,0),line(2.17,4,2.17,0),line(2.24,4,2.24,0),line(2.24,4,2.24,0),
line(2.31,4,2.31,0),line(2.38,4,2.38,0),line(2.45,4,2.45,0),line(2.45,4,2.45,0),
line(2.52,4,2.52,0),line(2.59,4,2.59,0),line(2.66,4,2.66,0),line(2.66,4,2.66,0),
line(2.73,4,2.73,0),line(2.8,4,2.8,0),line(2.87,4,2.87,0),line(2.87,4,2.87,0),
line(2.94,4,2.94,0),line(3.01,4,3.01,0),line(3.08,4,3.08,0),line(3.08,4,3.08,0),
line(3.15,4,3.15,0),line(3.22,4,3.22,0),line(3.29,4,3.29,0),line(3.29,4,3.29,0),
line(3.36,4,3.36,0),line(3.43,4,3.43,0),line(3.5,4,3.5,0),line(3.5,4,3.5,0),
line(3.57,4,3.57,0),line(3.64,4,3.64,0),line(3.71,4,3.71,0),line(3.71,4,3.71,0),
line(3.78,4,3.78,0),line(3.85,4,3.85,0),

line(3.86,0,3.86,3.99777778),line(3.93,0,3.93,3.94333333),line(4.0,0,4.0,3.88888889),line(4.0,0,4.0,3.88888889),
line(4.07,0,4.07,3.83444444),line(4.14,0,4.14,3.78),line(4.21,0,4.21,3.72555556),line(4.21,0,4.21,3.72555556),
line(4.28,0,4.28,3.67111111),line(4.35,0,4.35,3.61666667),line(4.42,0,4.42,3.56222222),line(4.42,0,4.42,3.56222222),
line(4.49,0,4.49,3.50777778),line(4.56,0,4.56,3.45333333),line(4.63,0,4.63,3.39888889),line(4.63,0,4.63,3.39888889),
line(4.7,0,4.7,3.34444444),line(4.77,0,4.77,3.29),line(4.84,0,4.84,3.23555556),line(4.84,0,4.84,3.23555556),
line(4.91,0,4.91,3.18111111),line(4.98,0,4.98,3.12666667),line(5.05,0,5.05,3.07222222),line(5.05,0,5.05,3.07222222),
line(5.12,0,5.12,3.01777778),line(5.19,0,5.19,2.96333333),line(5.26,0,5.26,2.90888889),line(5.26,0,5.26,2.90888889),
line(5.33,0,5.33,2.85444444),line(5.4,0,5.4,2.8),line(5.47,0,5.47,2.74555556),line(5.47,0,5.47,2.74555556),
line(5.54,0,5.54,2.69111111),line(5.61,0,5.61,2.63666667),line(5.68,0,5.68,2.58222222),line(5.68,0,5.68,2.58222222),
line(5.75,0,5.75,2.52777778),line(5.82,0,5.82,2.47333333),line(5.89,0,5.89,2.41888889),line(5.89,0,5.89,2.41888889),
line(5.96,0,5.96,2.36444444),line(6.03,0,6.03,2.31),line(6.1,0,6.1,2.25555556),line(6.1,0,6.1,2.25555556),
line(6.17,0,6.17,2.20111111),line(6.24,0,6.24,2.14666667),line(6.31,0,6.31,2.09222222),line(6.31,0,6.31,2.09222222),
line(6.38,0,6.38,2.03777778),line(6.45,0,6.45,1.98333333),line(6.52,0,6.52,1.92888889),line(6.52,0,6.52,1.92888889),
line(6.59,0,6.59,1.87444444),line(6.66,0,6.66,1.82),line(6.73,0,6.73,1.76555556),line(6.73,0,6.73,1.76555556),
line(6.8,0,6.8,1.71111111),line(6.87,0,6.87,1.65666667),line(6.94,0,6.94,1.60222222),line(6.94,0,6.94,1.60222222)), line(0,-3,0,12),line(-3,0,12,0)
,graph(400,8800/27,-.9,9.9,-.9,7.9,(17/4)x+4,(17/4)x-119/4,(17/4)x-8,(17/4)x-20)
 )}}}<br>
It should be clear, even without those extra lines drawn in the figure, that the two corners of the feasibility region where lines with slopes of 17/4 just touch the feasibility region are at (0,4) and (7,0).<br>
Then, evaluating the objective function at those two points...<br>
(0,4): 17(0)-4(4)+61 = -16+61 = 45
(7,0): 17(7)-4(0)+61 = 119+61 = 180<br>
ANSWER: the minimum value of P is 45 at (0,4); the maximum value is 180 at (7,0)<br>