What is the maximum possible volume of a cylindrical can with no top that can be made from 27pi square inches?
The surface area of a cylinder = 2 pi r h add in the bottom of the can for 2 pi r h + pi r² = 27pi divide through by pi and solve for : 2rh + r² = 27 h = (27 - r²)/2r Volume of a cylinder = pi r³h substitute above for h: V = pi r³ (27-r²)/2r which simplifies to: V = (27pi/2)r² - (pi/2)r^4 I don't know what class you are in. If it's calculus, take the 1st derivative, set to zero, and solve for r. I get r = √27 you can take it from here. that's it! ;)
What is the maximum volume of a rectangular box with a surface area of 300 in squared?
Given : xy + 2xz + 2yz = 300 i.e. xy + 2z(x + y) = 300 so, z = (300 - xy) / [2(x + y)] V = xyz = xy(300 - xy) / [2(x + y)] (on substitution for z) Take partial derivatives and set them equal to zero (for critical points) : ∂V/∂x = y^2(300 - 2xy - x^2) / [2(x + y)^2] = 0, implies y^2(300 - 2xy - x^2) = 0, because (x + y) ≠ 0, implies 300 - 2xy - x^2 = 0, because y ≠ 0. Thus, 2xy = 300 - x^2 (Equation 1) ∂V/∂y = x^2(300 - 2xy - y^2) / [2(x + y)^2] = 0, implies x^2(300 - 2xy - y^2) = 0, because (x + y) ≠ 0, implies 300 - 2xy - y^2 = 0, because x ≠ 0. Thus, 2xy = 300 - y^2 (Equation 2) From Equations 1 & 2, we have : 300 - x^2 = 300 - y^2 Therefore, y = x (note that both must be positive) Substituting back into the equation for V gives : V = x^2(300 - x^2) / (4x) = 75x - x^3/4 Take the derivative wrt x and set it equal to zero : dV/dx = 75 - (3/4)x^2 = 0, implies (3/4)x^2 = 75, or x^2 = 100, so, x = 10. x = y, so, y = 10. Substitute to find z : z = (300 - 100) / [2(20)] = 5 Thus, maximum volume is 10*10*5 = 500 in^3 when x (width) = 10 in., y (length) = 10 in. and z (height) = 5 in.
Calculus Problem-Max and min?
We need to minimize the surface area. S = 2πr^2 + 2πrh V = πr^2h = 60 h = 60/(πr^2) S = 2πr^2 + 2πr(60/(πr^2)) = 2πr^2 + 120/r = S(r) To minimize, S'(r) = 4πr - 120/r^2 = 0 4πr = 120/r^2 4πr^3 = 120 r^3 = 30/π r = 9.549296586... ≈ 9.549 The base radius is approximately 9.549 in.
Calculus optimization help?
I'm completely lost when it comes to optimization. In my head there just isn't enough information and well, I guess I just don't know the steps. :/ Yes, I know how to Google, and found this site http://www.math.ucdavis.edu/~strohmer/courses/21A/samplefinalsol.pdf with the the same question and answer. I'm just not sure how they got it. Anyway, this is the problem I'm stuck on: What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 1000 cm^3? The site said the answer is 10/(pi^(1/3)), but again, I have no idea why they did what they did to get that answer. It doesn't even sound like it answered the question which asks for dimensions and aren't dimensions something like 5 x 5? Very confused right now. Thanks in advanced!
A solid cube and a solid sphere of identical material and masses are heated to the same temperature and left to cool. which will cool faster?
Convective Cooling rate will be given by Newton's law of cooling,Q=hA(T-T`)& Radiation heat transferQ=kA(T^4-T’^4)k is Stefan Boltzmann constantUnder identical environmental conditions and temperature differenceheat transfer coefficient,h will be same.For same volumeSurface area of a sphere is less than a cube.So,Q will be more in case of a cube and it cools faster than sphere.( In practice,Nusselt Number is different for different body shape so 'h' will also be different)
When sugar is dissolved in water, the volume of the solution is the same. Where has the sugar gone to?
For instance, the sugar is still in the container you used.Now, a bit of chemistry lesson (perhaps more physical chemistry since it is the course during which I learned this stuff).When you look at a mixture of two liquids or a solid-liquid mix (like this one) the most important thing to remember is that volumes are not necessarily additive, which means that mixing 1 L of a substance and 1 L of another doesn’t imply a 2L solution as a result. The final volume could be 2 L but, more commonly, it would be a little less or a little more than 2, depending on which and how many substances we are dealing with.For example (this is very classic) mixing water with ethanol gives a final mix with a volume a little lower than expected. The property to discuss here is know as partial molar volume (which accounts for the variation in volume compared to common sense).On the other hand, masses are always additive (mass cannot be created nor destroyed. So 1 g of sugar plus 100 g ow water give 101 g of solution).What happens exactly? We must begin by assuming that we work at constant room temperature (dissolving something in a solvent strongly depends on temperature).So when you add sugar to water, some of its molecules (of the sugar), with their pretty crystal shape, find a space between the molecules of water and they actually do not use extra space. So if you add just a small amount of sugar to water and stir it until it’s all gone, you will see that basically nothing changes and if so, it is barely measurable.Adding more sugar will cause the new molecules to struggle in finding a place for their own, and since we have already reached maximum concentration, extra molecules will just deposit on the floor, since they are more complex, so “heavier” that the water molecules.These extra sugar molecules, by creating a non-dissolved layer, increase water level and, therefore, water volume.
Optimization question for calculus?
see http://math.about.com/od/formulas/ss/sur... for formula"s. Area top = pi(r^2) Area bottom = pi(r^2) Area around side = 2pi r h Total Area = 2( pi(r^2) + 2pi r h Volume = pi(r^2)h Contraints: V=1000 cm^3 h>=4, diameter>=6, which means r>=3 1000 = pi(r^2)h 1000/pi =r^2 h 1000/pi r^2= h 318.31/r^2 =h A= 2pi(r^2) + 2pi r h A= 6.2832(r^2) + 6.2832r h Sub for h, since h=318.31/r^2 A= 6.2832(r^2) + 6.2832 r (318.31/r^2) A= 6.2832(r^2) + 2000/r Take derivative to find max A, A’ =0 A’ = 2(6.2832r) + -1(2000)/r^2 0= 12.56637 r-2000/r^2 2000/r^2= 12.56637 r 2000=12.56637 r^3 2000/12.56637=r^3 159.155=r^3 5.419= r Find h=318.31/r^2= 318.309/29.368=10.838 The diameter = 2r =10.8385 Likewise the height = 10.8385 A= 6.2832(5.419^2) + 6.2832(5.419) 10.838 184.527 + 369.054 =553.58 sf of tin Tin dimensions side: 369.054=34.05 x 10.8385 for side Tin dimensions top: (pi)(5.419)^2=92.264 sf top, Tin dimensions bottom: (pi)(5.419)^2=92.264 sf bottom
How to calculate the optimal dimensions for a penny boat made out of one 6x6 sheet of aluminum foil?
I know the best way to design it but the professor wants to see the exact math behind it. Im not sure what i have to do to calculate the best dimensions for the boat. Any help would be appreciated. Thanks.