Applications of Maxima and Minima?

Question

1.Minimize S = x + 15y with xy = 15 and both x and y > 0.


2.Minimize F = 3x2 + 3y2 with xy2 = 16


3.Maximize P = xyz with x + y = 60 and y + z = 60, and x, y, and z 0.


Please explain how to find the solution.

Answer 1

S = x + 15y = x + 15(15/x) = x + 225/x
dS/dx = 1 - 225/x^2
dS/dX = 0 = 1 - 225/x^2
x^2 = 225, x = 15. (reject x=-15) xy=15, so y = 1

F = 3x^2 + 3y^2 = 3x^2 + 3(16/x) = 3x^2 + 48/x
dF/dx = 6x - 48/x^2
dF/dx = 0 = 6x - 48/x^2
6x^3 = 48. x^3 = 8, x = 2 (reject x=-2) xy^2=16 so 2y^2=16, y=SQRT(8)

Answer 2

S= x+15y = x +15*15/x = x+225/x
dS/dx = 1 -225/x^2
Set to 0 getting x^2=225 --> x = 15, s0 y=1

F= 3x^2+ 3*16/x
dF/dx = 6x -48/x^2
Set to 0 getting 6x^3 = 48 --> x = 2, so y =2sqrt(2)

Since x+y = 60 and y+z = 60, then x+y = y+z , so x=z
So let y =x so that x=30, y= 30 and z = 30
So pxyz = 30^3 = 27,000 for max

Answer 3

do you mean x*y = 15? then treat this as a constraint

S = x + 15y + L*(xy-15)

dS/dx = 1 + L*y = 0
dS/dy = 15 + L*x = 0
dS/dL = xy-15 = 0
solving, y = -1/L, x = -15/L xy=15 imlies that

(-1/L)*(-15/L) = 15 => 15/L^2 = 15 so L = 1 or -1

since y,x>0 then L = -1 and x=15, y=1