The product of two numbers is 192. Find the numbers when the sum of the first and three times the
second is a maximum.
+128408 Let the first be x and the second be 192 /x
And we wish to maximize
x+ (3 ) ( 192/x)
x + 576x^(-1) take the derivative and set to 0
1 - 576x^-2 = 0
1 576 / x^2
x^2 =576
x = 24 or x = -24
The second derivative is 2*576/ x^3 this will be negative when x = -24 so this is a (relative) max
So the numbers are -24 and 192/-24 = -8
