1.) if F(X) = 3x^2 -12x +5 find the absolute minimum and maximum values in the interval [0,3]
dF/dx = 6x-12
Find where 6x -12 = 0
x = 12/6 = 2
evaluate F(2) = 3(2)^2-12(2)+5 = 12-24+5 = -7
Therefore the minimum value of F on the interval [0,3] is (2,-7)
evaluate F(0) = 5 ==> (0,5)
evalute F(3) = 3(3)^2-12(3)+5 = -4 ==> (3,-4)
with a minnimum value at (2,-7)

2.) if G(X) = x^3 -3x + 1 find the absolute minimum and maximum values in the the interval [0,3]
dG/dx = 3x^2-3
Find where 3x^2-3 = 0
x = +- sqrt3/3 = +- 1
But we are ask to evaluate the interval [0,3] so we discard x = -1
evalute G(1) = (1)^3-3(1) + 1 = 1-3+ 1 = -1
Therefore the minimum value of G on the interval [0,3] is (1.-1)
evaluate G(0) = 1
evaluate G(3) = 27-9+1= 19
with a minnimum value at (1,-1)

3.) A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sideds of the rectangle. How can he do this so as to minimize the cost of the fences?

Since we are dealing with a rectangle we know that the area of a rectangle is equal to x side * y side, so x*y = 1.5 mil so x*y = 10^6 *1.5

And so y = (10^6*1.5) /x
Since we are cutting the rectangle in half we have the following picture:

We can see that the parimiter of the rectangel is going to be 3x+2y
and we found that y = (10^6*1.5) /x
so we can subtitute y into 3x+3y = 3x+2*(10^6*1.5 /x) = 3x+3*10^6/x
finding the Derivative of 3x+3*10^6/x
we get that dy/dx = 3-3*10^6/(x)^2 = 3(x^2-10^6)/(x^2)
The minimum will be found when (X^2-10^6) = 0
which occures when x = 10^3
we evaluate x = 10^3 in the origional equation y= (10^6*1.5) /x
y = (10^6*1.5)/(10^3) = 1500
Therefore we found that in order to minimize the cost of the fence the farmer should build the fence to be 1000 by 1500. That all x sides should = 1000 ft. and all y sides should equal 1500 ft.