Given that the curved surface area of a circular cone is 100, find out the maxium volume of this cone without using calculus.(our teacher doesn't allow us to calculate this with calculus!)
211345
Yes this can be done without calculus.\nThe volume, V, of a circular cone is:\nV = 1/3 x pi x r^2 x h \nwhere r = radius of circular base, and h = cone height\nYou also need the formula for, S, the surface area of the cone:\nS = pi x r x Sqrt(r^2 + h^2) (excludes area of base)\nWe need to know the ratio h/r which will give us the maximum Volume for a given S.\nAlthough this can be solved using calculus, we can determine the desired ratio h/r from an elementary consideration of planar figures:\nIf we are given that a rectangle has a given perimeter, p, then it is easy to show that a square with side length p/4 encloses more area (i.e. p^2/16) than any other rectangle with perimeter p. Similarly, an equilateral triangle with perimeter, p, will enclose more area than any other triangle having perimeter p.\nNow if we imagine rotating the equilateral triangle around its perpendicular height h (i.e a bisector of one of the angles) we will generate a cone with height h and surface area S and this cone will enclose a greater volume than any other cone having the same surface area S. For an equilateral triangle the ratio of the height, h, to 1/2 the base (which will become r for the cone) is h/r = Sqrt(3) = 1.732 or h = 1.732r\nSince we know S = 100 and we have h as a function of r, we can determine both r and h from the formula for S:\nwe have:\n100 = pi x r xSqrt(r^2 + (1.73r)^2). Solving for r we have\nr = 3.99 and h= 6.91. We substitute these values into the Volume formula to obtain the maximum volume:\nV = i/3 x pi x (3.99)^2 x 6.91 = 115.2 cubic units.
466051
let the radius of the cylinder be x,  drawing a cylinder in a cone on the paper, we see 2 triangles - the cone & the triangle with it's base as the top flat surface of the cylinder. follow me?\n\nif u do, then let's assume the height of the second triangle to be y. y can be found in terms of x as:\n(since both triangles are similar)\n\nx:4 :: y:12  \ni.e. x/4 = y/12\n=> y=3x\nthus ht of cylinder = ht of cone - y\n=12-3x\n\nvolume of cylinder = pi*r2*h  (r-> radius, h-> height of cylinder)\nv= pi * x2 * (12-3x) = pi * (12x2 - 3x3)\n  (the second term being 12 *x*x - 3*x*x*x ....sorry, i don't have a math equation display software here)\n\npartialy differentiate above equation wth respect to  x, we then get\ndv/dx = pi * (24*x - 9*x*x)\n\nequating dv/dx to zero, (to find a value of x that makes the volume value either maxium or minimum. u can verify it's max by again differentiating dv/dx with respect to x, equating to zero & finding if the value solved for is positive or negative)\n\nanyways, we get\n\neliminating  pi,\n24*x - 9*x*x=0\n\n=> x=24/9 = 8/3= 2.67\n\nnow, simply substitute this value of x in the volume equatiion above to get\n v= pi * (12*2.67*2.67 - 3*2.67*2.67*2.67)\n= 89.372 or so.\n\n\nuh...i guess i only confused u up, hope it helped u atleast a bit.
572887
here goes again .. seems like im gona do the whole hw for u ;) to find the limit of the area that ur gona revolve around the x=-2 axis u have to draw the two equations again and the area enclosed is basicly the shape of a circle (not a perfect circle though) and its located right above the x-axis touching it from below and the center of it is on the y-axis .. now the limit is the intersection of these two euations y=x^2 and y=2-x^2 that is x^2= 2- x^2 ==> 2*x^2 = 2 ==> x^2=1 ==> | x | = 1 ==> x = +1 or x = -1 .. and thats ur interval limits
270108
Volume of a cone is one-third the area of the base (a circle with a radius of three) times the height.\n\nFor slant height, think Pythagorean Theorem. Think of a cross section of the cone and draw a line for the height (down the middle) and a line for one radius. You would have two legs of a right triangle, with the hypoteneuse being the slant height. \n\nFor lateral area (that's the surface area of the slanted surface, the formula is pi x radius x slant height. (Sometimes it's given as 1/2 the circumference of the base times the slant height, but that's really the same thing as the formula I gave you.)\n\nHope that helps.
500897
Imagine a plane cutting through the center of your cone and cylinder. You will have a triangle with base on the x axis and vertex on the y, and a rectangle with base on the x axis and top 2 vertices on the triangle.\n\nCall the width of the rectangle 2x (so from 0 to a bottom vertex is x).\n\nWhere is the y for this? Say the base vertices of the triangle are at (-r, 0) and (r, 0), and the top vertex is at (0, h). Then the line connecting them is y = -(h/r)x + h\n\nNow set up a function based on h to compute the area of the cylinder:\n\nf(x) = pi*x^2*y = pi*x^2*((-h/r)x + h)\n= pi*h(x^2 - x^3/r)\nTake the derivative and set = to 0:\n\npi*h(2x - 3x^2/r) = 0\n2x = 3x^2/r\n2 = 3x/r\nx = 2r/3.\ny = 1/3h\n\nNow compute the volumes:\ncone: 1/3pi*r^2*h = 2/3pi*r^2h\ncylinder: 4/9pi*r^2*1/3*h = 8/27pi*r^2h\nDivide: (8/27)/(2/3) = 4/9
598382
Using cosine rule,\n\na² = b² + c² - 2bc cos A,\n\nwhere A=30º , b and c are the sides of cone (i.e. 14cm)  and a is the diameter of the cone.\n\n\na² = 14² + 14² - 2(14)(14)(cos 30º)\n     =  392 - 392 cos 30º\n     =  392- 392 (0.8660)\n     = 392- 339.472\n     =  52.528 cm\n\nDiameter, a = 7.2476 cm
225741
the formula for the area of any regular shape is \n(pa)/2, so there is no multipling the base by 2 involved.\np=perimeter\na=apothem
604973
i think n=6..\nyes this can be done..if you know the formula..\nof volume and surface arae of a hypersphere and also the form of gamma fucntion..\nhope you can do it yourself..i saw these two links and based on the formula given for volume and surface ares and the gamma fucntion expression its evident that n=6.\nHope you can figure it out yourself.\nvolume= pi^3/6 (radius =1 here)
419758
I have seen this question sometime this weekend!\nYour question is more complete, though.\nV=pi.r^2h=1000\nS=2pi.r(r+h)\n\nExpress h in terms of r\nh=V/(pi.r^2)\n\nSubstitute in S\nS=2pi.r(r+V/(pi.r^2))\n\nDifferentiate S w.r.t. r,\ndS/dr=2*pi*(V/(pi.r^2+r-2V/(pi.r^2))\n\nEquating this to zero and solve for r:\nr=(V/(2pi))^(1/3) and a negative root\nFor V=1000,\nr=5.41926\nh=1000/(pi.r^2)=10.8385\n\nThe result is not surprising, as the dimensions resemble that of a cube, or a sphere, which have the minimum surface area for a given volume.
596132
from calculus\ndivide the sphere up into circular slices of width dx.\nthe disk at a distance of x from the origin has a radius of sqrt(r^2-x^2) and so an area of pi(r^2-x^2) and a volume of pi(r^2-x^2)dx\nsumming up the volumes of all the disks gives the volume of the whole sphere = int pi(r^2-x^2)dx from x=-r to r\n=pi.r^2.x-pi/3.x^3 from x=-r to r\n=(pi.r^3-pi/3.r^3)-(-pi.r^3--pi/3.r^3)\n=4/3.pi.r^3
575836
A cone is, as you correctly state, a surface of revolution.  You can think of it as the volume enclosed in three-dimensional space by rotating a straight line passing through the origin around the x-axis.\n\nNow, because the straight line passes through the origin and rotates all the way around the x-axis to form the surface of the cone, at some point during its revolution, it must be in the plane formed by the x and y axes.  When it is, we can write its equation in two dimensions like this:\n\ny = mx + c\n\nBecause the line passes through the origin, c=0, so the equation becomes:\n\ny = mx         (equation A)\n\nNow for the calculus.  \n\nConsider a very small segment along the line.  As the line rotates, the tiny segment sweeps out a nearly-cylindrical volume symmetrical about the x-axis.  As the length of the segment tends towards zero, the volume it sweeps out becomes closer and closer to a cylinder.\n\nLet the distance on the x-axis corresponding to the line segment be dx.  What is the volume of the cylinder swept out at a particular point on the x-axis?\n\nWell, the volume of a cylinder is given by:\n\nV = pi * r^2 * h\n\nwhere r is the radius of the cylinder and h the height.  We know that the height of the cylinder is dx.  What about its radius?  This is given by the y-coordinate of the point where the cylinder touches the line.  We can determine this from equation A above: the radius is mx.\n\nSo the volume (call it dV) of this very thin cylinder is given by\n\ndV = pi * (mx)^2 * dx\n\nTo find the volume of the whole cone, let its height, measured from tip to base along the axis of symmetry (the x-axis, remember), be H.  Now to find the total volume we need to add up all the different dV's along the x-axis from 0 to H.  In calculus, this is called integration and is normally represented by a sign like a big old-fashioned letter S.  I can't write that here, but here goes:\n\nV = Integral from 0 to H (dV)\n= Integral from 0 to H (pi * (mx)^2 * dx)\n= pi * m^2 * Integral from 0 to H (x^2 dx)\n= pi * m^2 * 1/3 * (H^3 - 0^3)\n= pi * m^2 * 1/3 * H^3\n\nTo find the volume of a truncated cone, instead of integrating from 0 to H, choose H1 and H2 so that H2>H1 and integrate between them, giving\n\nV(truncated) = pi * m^2 * 1/3 * (H2^3 - H1^3)\n\nIf your problem is stated in terms of the radii of the ends of the truncated cone and its height, choose m, H1 and H2 so that \n\nm*H1 = r1 = radius of smaller end\nm*H2 = r2 = radius of larger end\nH2-H1 = H = height of truncated cone\n\nThis gives you three simultaneous equations in three variables which you can easily solve:\n\nm*(H2-H1) = r2-r1\nm*H = r2-r1\nm = (r2-r1)/H\n\nm*H1 = r1\n((r2-r1)/H)*H1=r1\nH1 = r1*H/(r2-r1)\n\nsimilarly\n\nH2 = r1*H/(r2-r1)\n\nSo the volume is given by\n\nV = pi * m^2 * 1/3 * (H2^3 - H1^3)\n\nsubstituting in the solutions and simplifying, we get:\n\nV = 1/3 * pi * H * (r2^3-r1^3)/(r2-r1)\n\nAs a quick sanity check, let r1 be zero and r2 be R.  We get\n\nV = 1/3 * pi * H * R^2\n\nwhich from high school geometry we know is the formula for the volume of a cone with height H and base radius R.\n\nThis would have been easier to explain with diagrams, so I hope it makes sense!
173838
The area inside the outer curve, a circle of radius 3, is pi r^2 or 9pi.\n\nFrom that you need to subtract the area inside the inner curve, which is the double integral\n\nouter: theta from 0 to 2pi\ninner: r from 0 to 2 + cos(theta)\nof r dr d(theta)\n\nSo find f(theta) = integral for r from 0 to 2 + cos(theta) of r dr [as if theta were a constant] and then find the integral for theta from 0 to 2pi of f(theta) d(theta)
543312
The solution is too convoluted to put it here without diagrams and a good equation editor.  If you wish, I can email a Word document detailing the solution.  I figure that the cone is (1 + 2*sqrt(2/3) + sqrt(3))*r (about 4.365 r) high and has a base radius of ((sqrt(1/2) + sqrt(4/3) + sqrt(3/2))*r or about 3.087.
182403
Let R and H be the radius and altitude of the CONE\nLet r and h be the radius and altitude of the CYLINDER\nVc=Vol of Cylinder\n\nThen Vc = pi*r^2*h\n\nIf the radius of the cylinder is zero (r=0), its altitude is the height of the cone (h=H).\n\nIf the radius of the cylinder is the same than the radius of the cone (r=R), its altitude is zero (h=0).\n\nFrom the above conclude:\n\nh=H when r=0   and   h=0 when r=R\n\nTreat h as a dependent variable (y) and r as an independent variable (x). Then, the slope of the cone (from the line above) is:\n\n(H-0)/(0-R) = -H/R\n\nTherefore the altitude of the cylinder "h" (the dependent variable) as a function its radius "r" (the independent variable) is:\n\nh=H-(H/R)r\n\nPlug this "h" variable in the formula of Vc\n\nVc = pi*r^2(H-(H/R)r)\n\nCompute the derivative of this function with respect to r (the independent variable. Remember H and are unknown but constants) and equate to zero. Next solve for r.\n\nr=0 (would yield a minimum)\n\nr=(2/3)R is the maximum\n\nPlug this value in the formula of "h" given above to obtain:\n\nh=H/3\n\nPlug this two values of r and h in the original formula Vc and obtain YOUR ANSWER:\n\n(4/27)*pi*H*R^2\n\nCompare to the volume of the cone, a little less than half of it.
413904
Yes, the answer is h = 2r\n\nThis is a classic optimization problem. You want the minimum surface area(SA) for a cylindrical shaped tin for a fixed Volume(V)\n\n(1) SA = 2pi r^2 + 2pi r h\n\nV = pi r^2 h (for simplicity set V = 1)\n\nso (2)    h = 1/(pi r^2)\n\nSubstituting this value for h into (1) gives:\n\nSA = 2pi r^2 + 2/r\n\nd/dr SA = 4pi r - 2/r^2\n\nd/dr SA = 0 when 2/r^2 = 4pi r\n\nthat is: (3)   r = 1/(2 pi r^2)\n\nTo get the value of the ratio h/r divide equation (2) by equation (3), which yields a value of 2 to 1.\n\nh = 2r
284710
Here's the correct solution.\n\nA cylindrical package with height(length) h and radius r will provide the maximum volume under the restriction that the sum of the girth and length equals a given constant C.\n(In your case C = 108")\n\nGeneral case, for arbitrary C\nFor a cylinder, the girth = 2pi*r, so our restriction is:\n\nC = 2pi*r + h\n\nThe volume V = pi*r^2*h\n\nFirst solve for h in terms of C:\n\nh = C - 2pi*r  Substitute this into volume formula:\n\nGives  V = pi*(r^2)*(C-2pi*r)\n\nV = pi*C*(r^2) - 2(pi^2)r^3\n\ndetermine dV/dr = 2*pi*r*C - 6*(pi^2)r^2\n\nSet dV/dr = 0, implies 6*(pi^2)r^2 = 2*pi*r*C\n\nSo 2pi*r = 2C/3, and therefore h = C/3, since 2pi*r + h= C\n\nSo if C = 108" then for maximum volume h= 108/3 =36\nand 2pi*r = 2C/3 so r = C/3pi and r = 108/3pi = 11.46 \n\nMax Volume = pix(11.46^2)x36 = 14,853 cubic in.\n\nWe can test that this is a maximum volume by testing the neighborhood. for example 3C/10 < C/3 < 3C/8\n\nlet h=3C/8 = 40.5 then r = 67.5/2pi= 10.74\n\nand Volume = pi*(10.74^2)*40.5 = 14,684 cubic in. This is less than the Max volume of 14,853 \n\nIf h= 3x108/10 then Volume = pi(12.03^2)*32.4 = 14,730\n\nThis is again less than Max volume, and difference increases the more h varies from C/3 = 108/3 = 36
388980
Volume = pi * r^2 * h\n\nwhere\nr is the radius\nand\nh is the height\n\nFor example:\nr = 0.9m\nh = 1.2m\nvolume = 3.14159 * 0.9 * 0.9 * 1.2 = 3.0536 cubic metres or 3053.6 litres\n\n_