basically, my teacher gave us a bunch of optimization problems and i've been working on them for hours and can't get them. if i could have help with maybe the first four, that would be AWESOME. thanks.

Question

1) find the point on the graph of the function y = x^2 that is closest to the point (14, 1/3). (i tried using the distance between two points formula/the pythagorean theorem here, but couldn't figure it out!)

2) find the maximum volume of a cylinder under a parabola defined by the function y = 12 - x^2.

3) a can company needs to manufacture cans which hold 256 cm^3 of liquid. what are the radius and altitude that will use the least amount of aluminum?

4) you are trying to make an isosceles trapezoid with 60 degree and 120 degree angles. what is the maximum area the trapezoid can have using (exactly) 50 cm of perimeter?

this is just an optional hand-out to help us understand optimization, but i really want to complete it. thanks in advance!

Answer 1

D = √(x-14)^2 + (y-1/3)^2)
square both sides, before differentiating and replacing y with x^2.
D^2 = (x-14)^2 + (x^2-1/3)^2

2D(dD/dx) = 2(x-14) + 2(x^2 - 1/3)(2x)
we set dD/dx = 0 for a max/min of D

2x - 28 + 4x^3 - (4/3)x = 0
6x^3 + x - 42 = 0

I sent this through my favourite cubic equation solver
http://www.1728.com/cubic.htm
and got x = 1.88389
subbing that into y = x^2
gave me y = 3.549

so the closest point is appr. (1.88,3.55)

I find it odd that the cubic did not work out easier. Usually question of this type are designed to give more civilized solutions.

Answer 2

Your #2 question does not make any sense.
the volume of a cylinder is 3-dimensional,
a parabola such as y = 12 - x^2 is 2-dimensional.

are you sure you stated the question correctly?

Answer 3

#3
let the radius of the can be r cm
and the height h cm
we know pi(r^2)h = 256 or h = 256/(r^2pi)

Surface Area = 2pi(r^2) + 2pi(r)(h)
= 2pi(r^2) + 2pi(r)(256/(r^2pi)
= 2pi(r^2) + 512/r

s(SA)/dr = 4pi(r^2) - 512/r^2 = 0 for a max/min

solving this for r
I got r = 3.441
sub that back to get h
I got h = 6.882

(notice that for a max volume the height and the diameter are the same)

Answer 4

for #4, I will give you some hints
and let you finish it

let each of the equal sides be y
let the shorter of the parallel sides be x.

If you extend this shorter side by the length of y, you will be able to complete an equilateral triangle with sides y.
You now have a parallelogram, and it should be obvious that the longer side of the original trapezoid is x+y

recall that area of a trapezoid
= (sum of the two parallel sides)(height)/2

The height of the trapezoid can be found from the equilateral triangle using the 30-60-90º property.

Give it a try.

Answer 5

of course you will also have the equation
2x + 3y = 50 from the perimeter

Solve for y and sub it into the Area equation, simplify, then differentiate.