Quadratic Optimization Problem Set

1. What real value of t produces the smallest value of the quadratic t^2 -9t - 36

Since this is  an upwards-turning parabola, the "t" that produces the smallest possible value  is given by :

-b/ [2a] =  -[9] / [2 (1) ] =   9/2 = 4.5

2. If t is a real number, what is the maximum possible value of the expression -t^2 + 8t -4?

This is similar to (1).....this parabola turns downward.......so the t value that produces the max is:

-8 / [2(-1)] =   -8 / -2  =    4

Putting this value back into the function we have the max value of

-(4)^2 + 8(4) - 4 =   -16 + 32 - 4 =    12

4. Given that xy = 3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x+3y/5.

If xy = 3/2  then y = 3/[2x]

So we have

10x + (3/5)[3/(2x)]

10x + 9 / [10x]

10x + (9/10)x^-1      take the derivative  and set to 0 

10  - (9/10)x^-2   =  0       multiply through by x^2

10x^2 - (9/10)  = 0

10x^2  = 9/10     multiply throgh by 10

100x^2 = 9       divide though by 100

x^2  = 9/100     taking the positive root, x = 3/10

And taking the second derivative procuduces a positve result, so this is a minimum

And xy = 3/2.....so........3/10 * y   = 3/2  →   y /10 = 1/2  → y = 5

So {x, y} = { 3/10, 5}     produces a minimum ....where x,y > 0