Each of x and y is a real number that satisfies this equation:
2x + y = -6
Determine the minimum value of this expression:
(1/4)x + (1/2)y
First I am going to start with a shortcut. Let z=2x. Then the constraint equation becomes z + y = -6 and the expression to minimize becomes (1/2)^z + (1/2)^y. These are perfectly symmetrical with respect to z and y, so that suggests a local extrema at z=y. Then z=y=-3 and (1/2)^-3 + (1/2)^-3 = 16.
This may or may not be a global extreme. So now for some proper rigor.
I'm not going to substitute immediately, as Larry did. Instead I will start by differentiating both the constraint and expression (call it f) with respect to x.
Then we get 2 + dy/dx = 0 and df/dx = ln(1/4)*(1/4)^x + ln(1/2)*(1/2)^y*(dy/dx)
A bit of cleanup: dy/dx = -2 and df/dx: df/dx = 2*ln(1/2)*(1/2)^(2x) + ln(1/2)*(1/2)^y*(dy/dx)
df/dx must equal 0 for an extrema to occur so then 2*ln(1/2)*(1/2)^(2x) + ln(1/2)*(1/2)^y*(dy/dx) = 0
Simplify a bit: 2*(1/2)^(2x) + (1/2)^y*(dy/dx)=0
Isolate dy/dx: dy/dx = -2*(1/2)^(2x-y)
This expression for dy/dx must equal the value found earlier. So then we have -2 = -2*(1/2)^(2x-y). From which we conclude 2x-y=0.
Then 2x + y = -6 and 2x - y = 0 gives us a solution at x=-3/2 and y=-3, from which we get the same result of the value of the expression is 16 at its only extrema. A quick sketch of the graph shows it is indeed a minimum, as requested by the problem statement.
Another solution
