Step-by-step explanation:
in such problems with just equations we always need to have the main (optimization) equation or condition, and then usually some constraint equations.
the constant equations allow us typically to express one variable through another, until we have only one variable for the optimization equation. then we find the zeros of the first derivative of that functions and find the needed maximum or minimum if needed via the second derivative.
now if we have inequalities, it is even easier, I think.
we find the corners of the shaded area that must contain our minimum or maximum by finding the intersections of the delimiter lines. the optimum must be one of the corners.
and then we try the coordinates of the different corners in the optimization equation and see, which one delivers the highest or lowest value. and we are done.
a.
x + y <= 240
y >= 4x
b.
500x + 450y = profit
theoretically we have 3 corners to test
(0, 0)
(0, 240)
(48, 192)
(0, 0) is clearly the minimum, profit 0.
(0, 240) gives
450×240 = $108,000
(48, 192) gives
500×48 + 450×192 = 24000 + 86400 = $110,400
that is the maximum.
so, the farmer makes the most profit under the given constraints, if he plants 48 acres corn and 192 acres sprouts.
