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.
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!
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.
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?
#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)
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.
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.
Of course! I'll be happy to help you with these optimization problems. Let's go through each one step-by-step.
1) To find the point on the graph of the function y = x^2 that is closest to the point (14, 1/3), we need to minimize the distance between these two points. We can use the distance formula or the Pythagorean theorem, as you've mentioned.
Let's call the point on the graph of y = x^2 as (x, x^2), since any point on the graph can be represented this way. Then, the distance formula between two points (x, x^2) and (14, 1/3) is:
distance = sqrt((x - 14)^2 + (x^2 - 1/3)^2)
To find the point that minimizes the distance, we need to take the derivative of this expression, set it equal to zero, and solve for x. Then, substitute the x-value back into the original equation to find the y-value. Finally, you'll have the coordinates of the point on the graph that is closest to (14, 1/3).
2) To find the maximum volume of a cylinder under a parabola defined by the function y = 12 - x^2, we need to optimize the volume of the cylinder. The volume of a cylinder is given by V = πr^2h, where r is the radius and h is the height of the cylinder.
First, we need to express the radius, r, and the height, h, in terms of x (the variable along the x-axis). Since the parabola is defined by y = 12 - x^2, the radius would be (12 - x^2), and the height would be x (the distance from the x-axis to the parabola).
So, the volume V can be expressed as V = π(12 - x^2)^2 * x.
To find the maximum volume, we need to differentiate this expression with respect to x, set it equal to zero to find the critical points, and check for the maximum using the second derivative test. The x-value corresponding to the maximum volume will give us the radius and height that maximize the volume of the cylinder.
3) To find the radius and altitude that will use the least amount of aluminum to manufacture cans holding 256 cm^3 of liquid, we need to minimize the surface area of the can. The surface area of a cylindrical can is given by A = 2πrh + πr^2, where r is the radius and h is the height (altitude) of the can, and A is the surface area.
Given that the volume V is 256 cm^3, we have the equation V = πr^2h. Thus, h = V / (πr^2).
Substitute this value of h into the surface area equation to get A in terms of r only. Then, differentiate A with respect to r, set it equal to zero to find the critical point, and check for the minimum point using the second derivative test. The radius corresponding to the minimum surface area will give us the optimal radius, and we can substitute it back into the equation for h to find the optimal altitude.
4) To find the maximum area of an isosceles trapezoid with 60-degree and 120-degree angles, using exactly 50 cm of perimeter, we need to optimize the area of the trapezoid.
Let's assume the bottom base of the trapezoid (the longer base) has a length of x cm. Then, the other base (the shorter base) would also have a length of x cm because it is an isosceles trapezoid.
Since the total perimeter is 50 cm, the sum of the lengths of the legs (the two non-parallel sides) would be 50 - 2x cm.
To solve for the height (the distance between the two parallel bases), we can use the tangent of the 60-degree angle as it is the opposite side divided by the adjacent side, which is equal to the height divided by half the difference in base lengths.
Once we have the height in terms of x, we can then use the area formula for a trapezoid, which is A = (b1 + b2) * h / 2, where b1 and b2 are the lengths of the parallel bases and h is the height of the trapezoid in order to express the area A in terms of x.
To find the maximum area, differentiate the area expression with respect to x, set it equal to zero to find the critical point, and check for the maximum point using the second derivative test. The x-value corresponding to the maximum area will give us the base lengths of the trapezoid, and we can substitute it back into the height equation to find the optimal height.
I hope this explanation helps you understand how to approach these optimization problems. If you have any further questions or need clarification, please feel free to ask!