i. y'= 3x^2 - 12
ii. here we wanna find out where the tangents are horizontal soooo... we set it equal to zero!
3x^2 - 12 = 0
3x^2 = 12
x^2 = 4
x = +2, -2
so le'ts plug these guys back in2 the original equation:
(2)^3 - 12(2) + 7 = -9 --------> (2, -9)
(-2)^3 - 12(-2) + 7 = 23 ------> (-2, 23)
iii. here we wanna find out when the graph is decreasing and increasing soo we choose points!
let's try -1000000000000
3(-1000000000000)^2 - 12 = some positive number!!! so increasing!!!
let's try 0
3(0)^2 - 12 = some negative number!!! so decreasing!!!
let's try 1000000000000
3(1000000000000)^2 - 12 = some positive number!!! so increasing!!!
sooooo.... the graph is increasing from (-infinity, -2) U (2, infinity)
and decreasing from (-2, 2)
b) i. since u know this one i'll go hed and skip it!
ii. we know the total surface area = 2040
so let's come up w/ an equation for it!!!!!
we know the base and lit of the box are squares, so the surface area of a square is x^2 but we have two of these so let's go hed and multiply that by 2 so it looks like this 2x^2
if we cut open the box and spread out the pieces we will end up w/ 4 rectangles and 2 squares. so the surface area of a rectangle is x*(height) and we have four of these so let's multiply that by 4 so it looks like this 4xh.
now we have an equation!!!! surface area = 2x^2 + 4xh = 2040
oh noes!!!! but we wanna maximize the volume and only got the surface figured out... bummer.
we know the volume of the box = x^2 * h.
now lets solve for h in the surface euqation:
510/x - 1/2 * x = h
and plug that in in2 our volume equation:
volume = x^2 (510/x - 1/2 * x)
= 510x - 1/2 * x^3
iii) let's find the derivative of the volume!!!!
=510x - x^3 * 1/2
= 510 - 3/2 * x^2
now we wanna find out the maximum volume of the box, but how do we do that??? well, let's find out the largest point possible!!!! set the derivate of the volume equal to zero
510 - 3/2 x^2 = 0
x = +(340)^(1/2) or -(340)^(1/2)
but we know the volume can't be negative!!! silly!!! so let's only keep positive answers!!
finally, we must prove this is the largest value, so in order to do this, we wanna c how the graph behaves. let's find out where it deacreses and increases by choosing points
lets try 5
510 - 3/2 * (5)^2 = some positive number!!! increasing baby!!!!
lets try 2000000000000000
510 - 3/2 * (2000000000000000)^2 = some negative number!!! oh dear!!!
finally, we know the graph is increasing from zero to (340)^(1/2) and decreasing from (340)^(1/2) all the way to infinity land!!! so x = (340)^(1/2) must be the maximum point!!!