Graph all the boundary lines' equations using = signs for ≤

7x + 9y = 63   <--that's a slanted line with intercepts (0,7) and (9,0)
y = 0          <--that's the x-axis
y = 4          <--that's a horizontal line thru 4 on the y-axis 
x = 0          <--that's the y-axis   
x = 7          <--that's a vertical line through 7 on the x axis



Consider 0 ≤ y ≤ 4 as y ≥ 0 and y ≤ 4

Consider 0 ≤ x ≤ 7 as x ≥ 0 and x ≤ 7

We decide which sides of all these lines to shade:

7x + 9y ≤ 63   <--shade below the slanted line because (0,0) satisfies it
y ≥ 0          <--shade above the x-axis (y greater than 0)
y ≤ 4          <--shade below the horizontal line (less than 4) 
x ≥ 0          <--shade to the right of the y-axis (greater than 0)  
x ≤ 7          <--shade left of the vertical line (less than 7)



We find all the corner points of the feasible region.
We already have three of them, (0,0), (0,4), and (7,0).

We find the other two by 

substituting y=4 into the slanted line's equation and solving for x, and
substituting x=7 into the slanted line's equation and solving for y.

     7x + 9y​ = 63            7x + 9y = 63   
   7x + 9(4) = 63          7(7) + 9y = 63
     7x + 36 = 63            49 + 9y = 63 
          7x = 27                 9y = 14
           x = 27/7                y = 14/9

So the other two corner points are (27/7, 4) and (7, 14/9)



Now we find the maximum and minimum values by substituting each corner point
into P = 17x - 4y​ + 61.

Corner point    P = 17x - 4y + 61 = value
  (0,0)         P = 17(0)-4(0)+61 = 0-0+61 = 61
  (7,0)         P = 17(7)-4(0)+61 = 119-0+61 = 180
  (7,14/9)      P = 17(7)-4(14/9)+61 = 119-56/9+61 = 1564/9 = 173 7/9
  (27/4,4)      P = 17(27/4)-4(4)+61 = 459/4-16+61 = 639/4 = 159 3/4
  (0,4)         P = 17(0)-4(4)+61 = 0-16+61 = 45

So: 
The maximum value is P = 180 when x = 7 and y = 0.
The minimum value is P = 45 when x = 0 and y = 4.

Edwin