Answer:
x=33,y=11
Step-by-step explanation:
The first constraint implies x*y=363, while you are finding the minimum of x+3y. You can do this with multivariable calculus approaches, but I will assume the simplest method is desired and use single-variable calculus methods.
Since xy=363, y=363/x
Substitute this into x+3y, and you get:
f(x)=x+3(363/x)=x+1089/x
Since you have this function as a function of a single variable now, you can minimize it like any other single variable function.
f'(x)=1-1089x-2
by setting f'(x)=0, the solutions x=33 and x=-33 are found. x=-33 is an extraneous solution, since you are required to have positive numbers. So, you must investigate x=33 and confirm it is a minimum.
If x=33 is a minimum of f(x), then values less than x=33 will yield a negative value in f'(x), since the function f(x) must slope downward up until it reaches a minimum. Conversely, values greater than x=33 will yield positive values in f'(x).
1<33
f'(1)=-1088
-1088<0, so f'(1) is negative
40>33
f'(40)=511/1600
511/1600>0, so f'(40) is positive.
Since the function f(x)=1-1089/x2 has a critical point at x=33 and since values less than x=33 result in negative values of f'(x) and values greater than x=33 result in positive values of f'(x), x=33 is a local minimum of f(x).
