optimization calculus

Question

if x*y -48, determine the minimum value of x + y³ for positive values of x and y

Answer 1

x*y=48 ... y=48/x

Plug into x+y^3 ..... x+(48/x)^3=x+(48^3)/(x^3)

Now you say x+(48^3)/(x^3)=the function f(x)

Find the minimum of f(x) from there

hint: set the derivative=0

Answer 2

The key to min/max problems is to identifying the function you are minimizing, x + y³ , and any constraints on the variables: x,y > 0 and xy = 48 (I assume).

The constraint relation allows you to eliminate one of the variables in the function to be minimized:

y = 48/x and f(x) = x + (48/x)³

In order to find a critical point, we find when the 1st derivative (df/dx) is zero:

df/dx = 1 + 48³(-3)/x⁴ = 0 for x = ((3)(48³))^1/4 which simplifies to ((3⁴)(2¹²))^1/4 = 24

The second derivative will be > 0 as the first term drops out and the second term will be >0 --> a minimum ( positive curvature is a smiley face.)

You have to check the endpoints to ascertain whether you have found a global minimum (0 and infinity in this case, where the f(x) blows up positively) . They are not minima.

I'll leave it to you to find out what the value of the function is.