+24 1. What real value of t produces the smallest value of the quadratic t^2 -9t - 36?
2. If t is a real number, what is the maximum possible value of the expression -t^2 + 8t -4?
3. The temperature of a point (x, y) in the plane is given by the expression x^2 + y^2 - 4x + 2y. What is the temperature of the coldest point in the plane?
4. Given that xy = 3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x+3y/5.
+130070 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
+130070 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
+130070 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
