## Review for Final Exam—Part 2

1. Consider a rectangle of perimeter 12 inches. Form a cylinder by revolving this rectangle about one of its edges. What dimensions of the rectangle will result in a cylinder of maximum volume?
2. What angle between two edges of length 3 will result in an isosceles triangle with the largest area?
3. A fish is reeled in at a rate of 1 foot per second from a point 10 feet above the water. At what rate is the angle between the line and the water changing when there is a total of 25 feet of line out?
4. A large balloon is rising at the rate of 20 ft/sec. The balloon is 10 ft above the ground at the point in time that the back end of a car is directly below the bottom of the balloon. The car is traveling at 40 ft/sec. What is the rate of change of the distance between the bottom of the balloon and the point on the ground directly below the back of the car one second after the back of the car is directly below the balloon?
5. Find the rate of change of the radius of a sphere at the point in time when the radius is 6 feet if the volume is increasing at the rate of 8$\pi$ cubic feet per second.
6. Find the rate of change of the volume of a cylinder when its radius is 6 feet, if its height is always (3/2)-times its radius and its radius is increasing at the rate of 2 feet per minute.
7. A sheet of cardboard 3 ft by 4 ft will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume?
8. Consider all triangles formed by lines passing through the point $(8/9, 3)$ and both the $x-$ and $y-$axes. Find the dimensions of the triangle with the shortest hypotenuse.
9. A cylindrical can is to hold 20$\pi$ m3 The material for the top and bottom costs $10/m2 and material for the side costs$8/m2. Find the radius $r$ and height $h$ of the most economical can.
10. You are standing at the edge of a slow-moving river which is one mile wide and wish to return to your campground on the opposite side of the river. You can swim at 2 mph and walk at 3 mph. You must first swim across the river to any point on the opposite bank. From there, walk to the campground, which is one mile from the point directly across the river from where you start your swim. What route will take the least amount of time?
1. April 23, 2013 at 9:41 pm

Does anyone know how to do question 1? This seems to be an optimization problem. What is given to me is that the perimeter of the rectangle is p=12inches. From the perimeter equation, I know that 12=2x+2y (x being the length and y the width). Simplifying, I get the equation to be 6=x+y. I then solved for y which equals…y=6-x. Since we are asked for the max volume of the cylinder the rectangle produces, I know that the volume equation is v=pi*r^2*h. i dont know how to link the two equations so that i can continue with the problem.

• May 1, 2013 at 6:22 pm

@Cameron

The width of the rectangle is the circumference of the cylinder.

Circumference = 2*Pi*radius. -> y = 2*Pi*r;

The height of the cylinder is the length of the rectange.

x = h;

Solve for r in y = 2*Pi*r, and substitute the r value and the h in your Volume of a cylinder equation. After you have the new Volume formula, you have the two functions related. Then optimize the formula. **Remember to check to see if it is a max or min to see if it is the right number!!!

2. May 1, 2013 at 6:08 pm

so the perimeter equation is P=2x+2y or P=2h+2r and we know that P=12 so 12=2h+2r. so solving for h would make h=6-r. The volume equation would be h*pi*r^2. Then plug in for h would be v=(6-r)*pi*r^2 after expanding…you get v=pi(6*r^2-r^3) then take the derivative….which it would be pi(12r-3r^2) then set to 0 which you would have 12r-3r^2=0 then 3r(4-r)=0 which would be r=4, 0 but zero is not an r but 4 is the definite r. then using the equation from the beginning 12=2h+2r you plug in 4 for the 4 and you will get that the h=2..using the volume equation plug everything in V=(2)(pi)(16) the v=32pi which is the max volume!!! hope this helps

3. May 1, 2013 at 11:54 pm

help with question 10.
So I drew out the river horizontally and labeled where you are, the camp is, the width of the river (1 mile wide), and the distance of the side length (1 mile wide). From here, I drew a line straight across the river as if you were to swim directly across and then walk the whole length on the otherside of the river down to the camp sight. I also drew a diagonal line from where you started to someplace on the otherside of the riverbank (i called this point “P”). this made a triangle in which i labeled my hypotenuse “y”, the width of the river I kept as 1 since its not changing, and the last side of the triangle (the portion on the otherside of the river from where you are) “x”. the other part I know is since the whole distance from you to the camp site is 1 mile long, the distance of the WHOLE side length can be expressed as x+(1-x). This will account for the side length that apart of the triangle and the remaining distance to the camp site.

Now for the harder part that I need help on. I know that this problem has something to do with the pythagorean theorem and since the problem is involving 2 different rates ( a walking rate of 3mph and a swim rate of 2mph) we need to express it as…
2a*da/dt+ 2b*db/dt= 2c dc/dt. and now im lost. I also know that the width is not going to change so we can plug that value of “1” into the pythagorean theorem before we take the derivative so the equation will look somewhat different. Can someone give me steps from here?

Dont forget that if you have any questions regarding these practice exams to post them. thanks!

• May 3, 2013 at 5:28 pm

If you use the square root of (x^2+1) as the length of the hypotenuse, then you can plug it into the equation distance=rate/time which can be reformatted to be time=distance/rate. Then add the two portions of the problem together because the time it takes to swim + the time it takes to walk will equal the total time. After this then you find the derivative of the problem and set the equation equal to 0 and solve for x.

• May 3, 2013 at 6:15 pm

From the equation you gave it looks like you are trying to solve it as a Related Rates problem which it is not. It’s Optimization and in order to set an equation you need to have : a pythag triangle yes but the equation for this is 1^2+x^2=z^2 which is z=sqrt(1+x^2). Then what you need to do is set up an equation in order to derive it to find the shortest route(optimized route) which is set up as: f(x)= ((sqrt(1+x^2))/2)+((1-x)/3) The 1-x comes from the distance that you walk after you are done swimming(since the whole side is 1 and you already have an x at the bottom for your triangle, the other side is 1-x) and the divided by 2 and divided by 3 comes from the speed at which you are walking/swimming. Now then you take the derivative of this equation and solve by setting equal to 0 and solve for your x to be able to plug etc. Hope this helps! I am still working on taking the derivative of the equation to find x so if anyone has the answer for #10 and how you completed it let me know!

• May 3, 2013 at 7:47 pm

i have completed #10
using David’s f(x)= ((sqrt(1+x^2))/2)+((1-x)/3)
f'(x)= something really ugly on paper without simplyfing that we all can do
the critical point i found from setting the f'(x)= 0 was 2/sqrt(5)

he other two critical points i tested as candidates were 0&1 from the domain
f(2/sqrt(5))=.706
f(1)=.707
f(0)=.833

the question asks what rout will take the least amount of time so you see that f(2/(sqrt(5)) is the minimum value which means…

you would swim a distance of sqrt(1+(4/5)) to the other side of the river, – the hypotenuse of the triangle calculated using 2/sqrt(5) and 1 as the two legs via Pythagorean theorem

then you would walk the remaining distance of 1-(2/sqrt(5)) to the campground

4. May 2, 2013 at 12:00 am

Could someone please give me a hint as to where to start with number 7? I have recognized that it is an optimization problem and my final goal should be to find a maximum.

I was thinking we could call the amount cut off from the corners, x because each side of a square is the same length.. this would make my equation
Area of the Regtangle = (Length)(width)(height)=(3-2x)(4-2x)(x)=max x[0,3]

this is the approach i have started to take, but I would like to make sure that there is not another method to this before I continue… any suggestions?

• May 2, 2013 at 2:29 pm

@kirsten
I approached it the same way…i don’t think there is another method that i know of..just take the derivative of that equation and then solve!

• May 3, 2013 at 5:31 pm

You can simplify the equation and multiply all of the answers together before you take the derivative so you dont have to use the product rule. Also, the max that x can equal is 1.5 because you cant cut off the whole side of 3 feet and still have a box.

• May 3, 2013 at 7:27 pm

thanks Katie
and Jenna i didn’t even think of that but it makes complete sense! thank you for correcting me!

5. May 3, 2013 at 9:40 pm

for number two, ive drawn my picture so that its two right triangles with a hypotenuse of 3 and a leg of x and a height of sqrt(x^2-9)
so the total area of the isosceles triangle:
2(1/2)(x)(sqrt(x^2-9)
And the angle between the two hypotenuses is 2(sin(x/2))

with these two equations would you find the maximum value for x in the area equation and then plug that max into the angle equation for your answer?

i really cant think of another way to do this problem, but i don’t remember doing an example like this in class, i could be wrong though.

• May 3, 2013 at 9:45 pm

/l\
/ l \
3 / l \
/ l \ 3
/ l \
_/____l____ _\
x x
essentially this is how i drew my picture

• May 3, 2013 at 10:28 pm

I used the same picture to start and then re drew it after I found the new angle. I used trig to find the variables that I used to identify the sides that I didn’t have a value for in terms of theta. Then plug the replacements into the equation for area and find the derivative using the product rule and the chain rule. Set the equation equal to 0 and solve.

6. May 3, 2013 at 10:16 pm

Kirsten-
I dont know what that triangle is, but this is how I solved question 2.
Step 1: draw a basic triangle. Label the base “x” and the height “y”. The tip top angle of the triangle will be your theta. label the left and right side lengths with length 3. let the angle we are considering be the top angle (which is between both the side lengths of 3 and is opposite from the base(x)).
The question asks for the angle that maximizes the area of this triangle. So we know eventually we will be taking the derivative of the Area equation. the reason we do step 1 is because the area equation will now be in terms of base x height or (xy).

Step 2: draw a line from the tip top angle to the base. this will essentially cut the triangle into two smaller triangles and your angle aswell. From this step, your angle (theta) is now (theta/2) for both small triangles and the base is now x/2.
the reason for doing the second step is we can now take the sin and cos of the angle (theta/2).

Step3: take the sin and cos of the angle. solve sin for x and cos for y. your 2 equations you get from this step will substitute in for xy=area. x=6sin(theta/2) and y=3cos(theta/2).
The reason for this step is because we want our final equation that we take the derivative of to be in terms of theta (da/dtheta=x’y+xy’).

Step4: substitute values of x and y into the equation area=1/2 base x height. By substitution, our equation looks like area=1/2(6sin(theta/2))(3cos(theta/2))…..SIMPLIFY…..9sin(theta/2)cos(theta/2).

Step5: we are now finally ready to take the derivative. da/dtheta= CHAIN RULE.
Step6: set equation =0 and solve for theta by placing cosine and sine on opposite sides of the = sign, take the square root of both sides (if you did the chain rule right in step5), your cosine will just be cos(theta/2) but your sine will be (+/-)sin(theta/2).

Step7: evaluate the unit circle and you should get theta is equal to pi/2 OR 3pi/2.

Step8: create your intervals for a possible max or min. the intervals that theta could be in is from 0 <(r equal to) theta <(or equal to) pi.
By evaluating your results and the interval, you notice that 3pi/2 is outside the interval and therefore cannot be an answer.

Step9: set up a "sign line" and test all CRITICAL POINTS (0 pi/2 pi). from here you see that pi/2 is your angle that will get the max area for the triangle.

hope this helps.

7. May 3, 2013 at 10:39 pm

Hey, I was working on #5. I listed my variables down as:

V= 4/3*pi*r^3
dv/dt = 8*pi ft^3/s
r = 6ft
dr/dt = Unknown

I took the derivative of both sides of the equation and got:

dv/dt = (3*4*pi*r^2)/(3) * dr/dt

I simplified everything down and plugged in the variables. I ended up getting 8/144 ft/s, is this the correct answer? If not may someone help me?

• May 4, 2013 at 5:42 am

Yes Glenn, you did the same process that I did and we got the same answer, good job!

8. May 3, 2013 at 11:42 pm

Glenn i got the same answer !

• May 4, 2013 at 12:59 am

Ahh, great! Looks like we both got it right, (or hopefully we both didn’t get it wrong hahah).

9. May 4, 2013 at 6:56 am

I am working on number 4 and I have the picture as a big triangle. I have one side equal to 30 (20+10) and the other equal to 40. I used the pythagorean theorem to find the hypotenuse and did all the calculations and I got my final answer as 1800/sq(2500) is this the answer that anyone else got? If not, any help would be greatly appreciated!