<p>Q1:
A closed, cylindrical can is to have a volume of V cubic units. show that the can of minimum serface area is achieved when the height is equal to the diameter of the base.</p>
<p>does someone who can help me solve this prob?
Thx dude</p>
<p>(pi/4) x h^3?</p>
<p>the minimum SA is when h = d, so r = .5h</p>
<p>Volume of cylinder is (pi) x r^2 x h</p>
<p>plug .5h for r and solve.</p>
<p>THANK U
could u show more detail about the prob
do I need take the derivative?</p>
<p>Yes, the problem can only be rigorously solved with calculus (you didn't come across this question during SAT prep, did you?). It's an optimization problem (that term should be familiar to you if you are in calculus right now).</p>
<p>You have V=(pi)r^2(h) and SA=2(pi)r^2 + 2pi(r)h
You want to minimize SA, so pick an arbitrary number for the cylinder's volume. Say, 100. So, you have 100 = pi(r^2)h
Solve that equation for either h or r (I choose h). Then, h = 100/pi(r^2).
Plug your new value for h into your SA equation. You now have a 1-variable equation. Take the derivative with respect to h and set your new equation equal to 0. Find out what h equals. Plug your value for h back into your volume equation. Get the value of r. h should be twice the value of r.</p>
<p>Thank you so much
I am a sophomore right now and Im gonna take ap cal next term</p>
<p>i didnt use calc. in the problem, it tells you that the minimum SA is when h = d and the formula for the volume of a cylinder is (pi)r^2 x h. when you convert d to r, you get r = .5d and since d=h substitute h for d. you get r = .5h. plug in .5h for r and solve. you should get
(pi/4) x h^3.</p>
<p>Thxxxxxxx :)</p>
<p>
[quote]
i didnt use calc. in the problem, it tells you that the minimum SA is when h = d and the formula for the volume of a cylinder is (pi)r^2 x h. when you convert d to r, you get r = .5d and since d=h substitute h for d. you get r = .5h. plug in .5h for r and solve. you should get
(pi/4) x h^3.
[/quote]
No, the problem is asking you to SHOW that the minimum SA is achieved when h=d. Therefore, you cannot assume that h=d to solve the problem. You have to do the problem without assuming that, and h=d should be the answer of the problem.</p>
<p>actually you can assume that h=d since that is the only information given to you. it clearly states that the minimum surface area is achieved when the height is equal to the diameter of the base. what they mean by show is to find the volume in terms of h.</p>
<p>No, Latency is right. Just because they tell you what you have to show doesn't mean you can assume it. Doing that in regular life would sounds like something straight out of Alice and Wonderland.</p>
<p>Example:</p>
<p>Me: The moon's made of cheese.
You: Prove it.
Me: Prove what?
You: Show that the moon's made of cheese.
Me: Since the moon's made of cheese, it tastes delicious. Ergo, it must have some cheese in it. QED.
You: ...</p>
<p>I think Latency will agree that this is <em>exactly</em> what just happened.</p>
<p>yep I agree with that</p>
<p>HERE'S ANOTHER ONE:</p>
<p>(a) find the area of the region enclosed by the parabola y=2x-x^2 and the x-axis.
(b) find the value of m so that the line y=mx divide the region in part (a) into two regions of equal area. (show work)</p>
<p>I only can do part A but I cant do part B..
u guys who wanna chanllenge yourself?</p>
<p>You could try the following: find the x value (other than zero) at which the line and parabola intersect. Then find the two areas using integration, set them equal, and solve for m.</p>
<p>This is clearly not SAT stuff.</p>
<p>r u sure? I foud some questions like these problems on math part. Therefore I asked my bro to slove these problems but he couldn't do them</p>
<p>Definitely not SAT I math. No way.</p>
<p>do u know how to slove Q2?</p>
<p>Haha, AuburnMathTutor, yeah, right on the dot.</p>
<p>For those other two problems...you can't even solve the first one unless it gives you a range of values in which you want to solve the area for. If x is any real number, the area between the parabola and the x-axis is infinite since the parabola keeps expanding as x gets larger (or smaller). And even if it gave you a range of values, only integration can solve it.</p>
<p>As for the second question, wouldn't the line of symmetry down the middle of the parabola divide it into two equal areas? x=1? Thus m equals 0...</p>
<p>I'd like to know where you are getting these questions because they definitely aren't showing up in any legit resource related to the SAT. Maybe your just trying to get help with problems in your math class but you're trying to cover it up..</p>
<p>Well Latency, I might have to disagree with you on that.</p>
<p>I think he said the "area bounded by the parabola and the x-axis". To me, this means the finite area. And if</p>
<p>y = 2x - x^2, then y = x (2 - x) and y = 0 when x = 0 and when x = 2. Since the thing is concave down, it's maximum is (by inspection) at (1, 1) and therefore there is a finite area equal t 4/3 bounded by it (see below)</p>
<p>S y dx = x^2 - 1/3 x^3, plugging in from 0 to 2 you get: A = 4/3.</p>
<p>Now, if you want to split this area in two, each of the two areas will be 2/3. If the line must be:</p>
<p>y = mx</p>
<p>Then you can find where it intersects the parabola as a function of m:</p>
<p>mx = 2x - x^2, (m-2)x + x^2 = 0, x (m-2+x) = 0, x = 0 OR x = 2 - m.</p>
<p>Now you can easily find the area of the upper half; it's just going to be</p>
<p>S (y - mx) dx from 0 to 2 - m. This yields:</p>
<p>x^2 - 1/3 x^3 + 1/2 mx^2 = 1/2 (2 - m) x^2 - 1/3 x^3. Plugging in for the x's yields:</p>
<p>1/2 (2-m)(2-m)^2 - 1/3 (2-m)^3 = 1/6 (2-m)^3. Setting equal to 2/3 yields:</p>
<p>1/6 (2-m)^3 = 2/3 => (2-m)^3 = 4 => 2-m = 4^(1/3) => m = 2 - 4^(1/3).</p>
<p>This can be more neatly written as 2(1 - 2^(-1/3)) which equals, approximately, 0.4126.</p>
<p><em>checking for reasonableness with a graphing package</em></p>
<p>Looks about right.</p>
<p>So is that the right answer?</p>
<p>Oh, right. I'm in BC Calc right now doing integrals, and we've been doing a lot of problems where we're finding the area between the curve and the x axis, so I guess I jumped to a conclusion to interpret the problem as such. Obviously, the enclosed area just means the finite area above the x axis.</p>
<p>Anyway, that looks right. Integrate from 0 to 2 to find the area enclosed. Then find where the line intersects the parabola and integrate between those two points to find the area under that line. Then set the area under that line to half of the area to get m. Looks good.</p>
<p>Actually I'm a little confused as to what you did right here.
[quote]
This yields:</p>
<p>x^2 - 1/3 x^3 + 1/2 mx^2 = 1/2 (2 - m) x^2 - 1/3 x^3. Plugging in for the x's yields:
[/quote]
I see that you took the anti-derivative of y-mx (shouldn't it be - 1/2 mx^2?). But then, how did you set that equal to what you have on the other side of the equation?</p>