The first few are all similar.....finding mins or maxes on parabolas
In the form ax^2 + bx +c, the min [or max] will occur at the x coordinate =
-b / [2a] ..... and we will .plug this value back into the equation to find the minimum or maximum y value
t^2 -9t - 36
This is a parabola. The smallest value will be produced when t =
-(-9) / [ 2 * 1 ] = 9/2
This is the x value of the vertex....and on this parabola,it is the minimum point
If $s$ is a real number, then what is the smallest possible value of $2s^2 - 8s + 19$?
Similar to the first one.........the smallest value will occur when s = -[8] / [ -2 *2] = 8/4 = 2
And the minimum value is 2(2)^2 - 8(2) + 19 = 8 - 16 + 19 = 11
If $t$ is a real number, what is the maximum possible value of the expression $-t^2 + 8t -4$?
Max will occur at t = - 8 / [ 2 * -1] = -8 / -2 = 4
And the max value will be -(4)^2 + 8(4) - 4 = -16 + 32 - 4 = 12
The Art of Problem Solving has begun selling a cookbook called "What Would Euler Eat?" If the price of the cookbook is $n$ dollars ($n \le 72$), then it will sell $720 - 10n$ copies. What price (in dollars) will maximize the total revenue we receive for the books?
This one is a little tricky...... price * quantity = total revenue.....so our function is
n * (720 - 10n) = 720n - 10n^2 = -10n^2 + 720n
The price that maximizes the revenue will be = - [ 720] / [ 2 * -10] = $ 36
And the max revenue is -10(36)^2 + 720(36) = $12960
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?
Firstly...........this is a 3D object not capable of being represented in a 2D plane
We would need Calculus to solve this....the process is long and tedious.....anyway.....the minimum = -5 and it occurs at (2, -1)
Given that xy=3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x+(3y/5)
Rearrange the first equation as y = 3 / [ 2x]
Putting this into the second equation, we have........ 10x + 9 / [ 10x ]
We can rewrite this as 10x + (9/10)x^-1
Take the derivative and setting it to 0.....we get....... 10 - (9/10)x^-2 = 0
Multiply through by 10 and rearrange as
100x^2 = 9 divide both sides by 100
x^2 = 9/100 take the positive root = 3/10
So....the minimum is 10(3/10) + 9 / 3 = 6 at {x,y} = {.3, 5 }
I don't understand the last problem.......sorry....!!!
