what is the slant height?\nwhat is the lateral area?\nTotal area?\n&\nVolume?
132745
slant height = 12.08\ntotal surface area= 251.428\nvolume = 287.98
3363
Find the dimensions of the base:\n36 - 4 - 4 = 28\n25 - 4 - 4 = 17\n\nSo the area of the base = 28 x 17 = 476\n\nThe height = 4\n\nSo the volume = 4 x 476 = 1904
250970
I am assuming you mean something like a round flower pot, where the base is smaller than the top.  You are saying the bottom has a diameter of 17 inches and the top has a diameter of 34 inches, with a height of 34 inches (not the slant).\n\nIs this correct?  I've also heard this called a truncated cone.  The formulas for volume is given below:\n\nHeight: h\nRadius of bases: r, R\nVolume: V\n\nV = Pi(R^2+rR+r^2)h/3\n\nSo plugging in your values (don't forget to cut the diameters in half to get radii):\n\nh = 34\nr = 8.5\nR = 17\n\nV = Pi(17^2+(8.5)(17)+8.5^2) 34 / 3\nV = Pi(289 + 144.5 + 72.25) 34/3\nV = Pi(505.75)34/3\nV = Pi(505.74)(10.333)\nV = Pi(5731.833)\nV = 18007 cubic inches
206530
Surface area:\nCylinder(lateral):2 times pie times radius times height\ncylinder(total): 2 times pie times radiuse times height + 2 times pie radius squared\nTriangular prism(total): find the area of every side and add all together to get surface area(hint:area of a triangle=1/2bh or bh/2)\nCone(lateral):pie times radius times length\nCone(total):pie times radius times length + pie squared\npyramid(total):find area of ever side and add all together \narea of a triangle=1/2bh bottom of the pyramid is length times width
478296
I'll leave you to do some of it.\n\nLet the length of the base be x cm, the height h cm.  \n\nThe total area of base and sides is\nx^2 + 4*h, and since we are told this is 1200 sq cm, we write\nx^2 + 4h = 1200, which allows us to express h in terms of x.  [You can do that.]\n\nNow volume\nV = (x^2)*h, and if you replace h by the expression you just found, you have V as a function of x only.  I expect you've already done the mechanics of differentiation and finding max val, so I'll leave you to finish it yourself or copy from others who will post the complete working here or you can email  h_chalker@yahoo.com.au
624195
area or volume???\n\narea = 2 (3.4 * 3 + 3*2.2 + 2.2*3.4 )\n in square meters.\n\nvolm = 3.4*3*2.2\nin cubic meters.\n\n\nI only edited bcoz ever1 is giving wrong ans for area. Correct area is 48.56 sq mtr. I dont understand how it can b 6.6\n\nvolm = 22.44 cubic mtr
407756
1) Rearrange so\ny = -(5/3)x + (2/3)\nThis means the perpendicular to this line (which the point must lie on) is of the form\ny = (3/5)x + n \nlet (a,b) = the point we require, so\nb = -(5/3)a + (2/3)   equation 1\nand -5 = -4(3/5) + n so n = -(13/5)   equation 2\nand b = a(3/5) + n  equation 3\nsubstitution of equations 1 and 2 into 3 give:\n-(5/3)a + (2/3) = a(3/5) - 13/5\nso a = 49/34\nand consequently b = -59/34\nso the point is (49/34, -59/34)\n\n2) let h = height of box and l = width of box = length of box (square base)\nso volume, v = hl^2\nand material area, a =l^2 + 4lh = 1300 cm^2\nso h = (1300-l^2)/(4l)\nso now v = (1300l - l^3)/4\nfor max volume \ndv/dl = 0 \nso dv/dl = (1300 - 3l^2)/4\n(1300 - 3l^2)/4 = 0 when l = sqrt(1300/3)\nwe take only the positive solution as the negative is meaningless in this problem.\nto check this is indeed a maximum value (as oppose to a minimum) we derive d^v/dl^2:\nd^v/dl^2 = -3l/2 so for l = sqrt(1300/3) this is always negative confirming this is a maxima.\nso now, v  = (1300l - l^3)/4\nsubstitute l = 20.817 cm\ngives volume, v = 4510.28 cm^3
134116
See previous answer--\n\nGives vertical height as rt27\n\nTherefore 1/2*6*rt27 is area of one side\n\nBase is 6*6 = 36
503268
The closer you get to a cube shape (i.e., width=length=height) the better.\n\n**************\n\nOoops. You're right, it's deceptive. While a cube maximizes the volume for a GIVEN area, in this problem we decrease the total area the more we cut from the sheet. So, you need to express the volume of the cube in terms of the heights of the sides.\n\nI did a quick re-calc, and decided that the squares (i.e, the height of the resulting sides) should be 1/3 m.\n\nGood question!
123590
Think of the first as the area of a flat round plate sweeping out the inside of the figure you get from the rotation.  What will be its area? πr² where the r is the y value of √(9 + X²).  That changing area will sweep out the volume from whatever the limits are (you don't say), so it's the integral of π(√(9 + X²))², or π(9 + X²).\n\nIn the second case, the thing sweeping out the volume is the surface area of a cylinder as its radius grows.  Thats the circumfrence of the base times the height.  The circumfrence is 2πr where r is the X value and the height is the difference between Y = x and Y = X/2.  So it's the integral from 0 to 2 of:\n\n   2πX(X-X/2).\n\nThinking of an area sweeping thru the volume helps.
455445
Total surface area of a tin can = A = 2pr^2 + 2prh\nVolume of a tin can = V = pr^2h = 1000 cm^3\nTherefore, h = 1000 / pr^2\n\nSubstituting this into the area equation gives :\nA =  2pr^2 + 2pr * 1000 / pr^2\n= 2pr^2 + 2000 / r\n\nNow take the derivative of A with respect to r :\ndA / dr = 4pr - 2000 / r^2\n\nSet this equal to zero to find the minimum :\n4pr - 2000 / r^2 = 0\nThis can be rearranged to show that : 1000 / pr^2 = 2r, which equals h, as shown above.\nContinuing to find r, we get : r = (500 / p) ^ (1/3), or approx. 5.42 cm.\n\nNow h = 2r = 2 * (500 / p) ^ (1/3), or approx. 10.84 cm.\n\nThus, ratio is h / r = 2r / r = 2\n\nI'm a bit confused about the second question, but I think what you want is a square sheet of aluminium, such that all 3 pieces of a can will fit, with the minimum of wastage. This would be a square sheet with the dimensions : (perimeter of can) by (height + 2 * radius).\n\nThus, Perimeter = Height + 2 * Radius, because the sides of a square are equal.\nTherefore, 2*pi*r = h + 2r,\nso, h = 2r(pi - 1)\n\nBut, V = pi*r^2*h = 1000, so h = 1000 / (pi*r^2)\n\nEquating the 2 values for h gives :\n2r(pi - 1) = 1000 / (pi*r^2)\nfrom which we get : r  = {500 / [pi(pi - 1)]} ^ (1/3)\n\nSo now we know that h = 2r(pi - 1), so that after substituting for r, we get : \nh = 2 * (pi - 1) ^ (2/3) * (500 / pi) ^ (1/3)\nAnd we find that the ratio h / r = 2r(pi - 1) / r = 2(pi - 1)\n\nWell, that was a bit of speculation, but even if I've got it wrong, then perhaps you can adapt it to your needs.\n\nEdit\nCan't help feeling I've misunderstood the second question the first time around, but on re-reading it many times, I still don't understand it.
344610
Write down all relevant variables, and units:\nV = volume in mm^3\nt = time in min\nx = edge length in mm\n\nWhat down what you are told:\ndx/dt = 0.49\nV = x^3\n\nWrite down what you want:\ndV/dt\n\nTo get dV/dt, we need dV/dx and dx/dt. The second one is 0.49, and the first one can be gotten by differentiating the volume equation: dV/dx = 3x^2. Substitute in x = 6.7, and the final answer is \n3*6.7^2 * 0.49 = 65.9883 mm^3 / min (decreasing)
181613
actually it's lateral area plus 2 times the base\n\nif lateral area= L and base= B, then total area is L+2B
547643
Area A = area of two ends  +  Curved surfaces.\n  A  =  2π r²  +  2π r h\n\nVolume V = Area of end  x  Heigh.\n  V  = π r²  *  h\n  V  = π r² h\n  h  =  V / π r²\n Now substitute for h into the area formula.\n\n  A  =  2π r²  +  2π r h\n  A  =  2π r²  +  2π r (V / π r²)\n  A  =  2π r²  +  2V / r\nThe volume of the can is 1000cm^3, so substitue in.\n  A  =  2π r²  +  2πV / r\n  A  =  2π r²  +  2000 / r\n\nTake first derivide and let it equat to zero (zero for max. or min. pt.).\n  A  =  2π r²  +  2000 / r\n  δA/δr  =  4π r  -  2000 / r²\n  4π r  -  2000 / r² = 0\n  4π r  =  2000 / r²\n  4π r^3  =  2000\n  r^3  =  2000/ 4π\n  r  =  (2000/ 4π)^1/3    (that is, cubed root).\n  r  =  5∙419 260 701\n  r  ≈  5∙42 cm.\n\nLet's use the r value to calculate h.\n  V  = π r²  *  h\n  1000  = π (5∙419 260 701)²  *  h\n  h  =  1000 /π (5∙419 260 701)²\n  h  =  10∙838 5214\n  h  ≈  10 51 cm.\n\nFor ratio:\n h/r = 10∙838 5214 / 5∙419 260 701\n h/r = 2/1   That is, a two to one ratio.\n\nFor queation two, follow the same proceedure after you set up your formulas.\nLike wise for question three.\n\nWith question two it's not clear what value you are taken for the length of the side of the square edge.  Because you are still using r, I'm assuming you are taken the largest size of a square that fits into the circle or radius r.  Based on this assumsion, the calculations are are follows:\n\nTo identify the length x of one side of the square of radius r:\nDraw two diameters of the circle at 90º to each other.  Where these diameters make contact with the circumference of the circle will be the corner points of the square.  The area of this square now consists of four triangles.  These triangles will have two sides of length r with the enclosed angle of 90º.  So the length of x will be:\n\n x² = r² + r²\n x² = 2 r²\n x = √(2 r²)   That is, length of one side of the square.\n\nArea A = area of two square ends + parameter of square * heigh.\nA = (2r² * 2) + [(√2)r * 4 * h]\nA =  4r²  +  4√(2)r h\n\nVolume V = square area x heigh.\nV =  2 r² * h\nV =  2 r² h\nh = V / 2 r²\nNow substitute for h into the area formula.\n\nA =  4r²  +  4√(2)r h\nA =  4r²  +  4√(2)r V / 2 r²\nA =  4r²  +  2√(2) V/r\nA =  4r²  +  2000√(2) r^-1     (Volume is 1000cm^3).\n\nTake first derivide and let it equat to zero (zero for max. or min. pt.).\nA =  4r²  +  2000√(2) r^-1\nδA/δr = 8r  -  2000√(2) r^-2\n 8r  -  2000√(2) r^-2  =  0\n 8r  =  2000√(2) r^-2\n 8r  =  [2000√(2)] /r²\n 8r^3  =  2000√(2)\n r^3  =  [2000√(2)]/ 8\n r  =  ([2000√(2)]/ 8)^1/3             (that is, cubed root).\n r  =  (353∙55 33 906...)^1/3\n r  =  7∙071 0678 12\n r  ≈  7∙07 cm.\n\nLet's use the r value to calculate h.\nV = 2r² * h\n1000 = 2r² * h\n1000 = 2 (50) * h\n1000 = 100h\n h = 10 cm.\n\nFor ratio:\nh/r = 10 / 7∙071 0678 12\nh/r = 1∙845 270 149 / 1\n h : r  ≈ 1∙85 : 1\nThese ratios do not take into account the scrap metal from the circle when the square is cut out of it.
641056
Area of rectangle = length x width \nArea of a circle = pi x R x R = pi R^2 where pi~3.14 and R - radius\nArea of an ellipse = pi x a x b see http://www.tpub.com/content/engineering/14069/css/14069_33.htm \n\nVolume = area of the base times hight\nVolume of a rectangular prism= area of rectangle x height = length x width x height\nVolume of a sphere = (4/3)pi x R x R x R = (4/3)pi x R^3\n \nCheck math tables or the web for equation of other geometric figures \n\nPercentage = 100 x (partial amount) / (total amount)\n\nHave fun
547326
1. A sphere with a diameter of 16 mm has the same surface area as the total surface area of a right cylinder with the base diameter equal to the sphere diameter. how high is the cylinder?\nA (sphere)=4*pi*r^2=4*pi*8^2=256*pi mm^2\nA cylinder=2*pi*r^2+2*pi*r*h==128*pi+16*pi*h\nA sphere=A cyl\n256*pi=128*pi+16*pi*h\n128*pi=16*pi*h\n8=h\nheight is 8 mm\n\n2. find the volume of a frustum of a right pyramid whose lower base is a square with a side 5 in. , whose upper base is a square with a side 3 in. , and whose altitude is 12 in.?\n\nB1 area=25in^2\nB2 area=9in^2\naltitude=3\nV=1/3(3)(25+15+9)=49in^2\n\n3. The figure formed by the intersection of a solid with a plane parallel to the base of the solid is congruent to the base if the solid is a........? \n\nprism
513642
Express volume as a function of depth by using the fact that the volume of the frustum is the difference between the volume of two cones and the fact that the ratio between the depth plus 6 inches and the radius at the surface is 12/2:\nV(x)=1/3*(x+6)*pi*((x+6)/6)^2-1/3*6*pi*1^2=pi/108*(x+6)^3-2pi.\nNow differentiate with respect to time, assuming that x is a function of time:\ndV/dt=pi/36*(x+6)^3*dx/dt.\nNow, remember that dV/dt=-3 and x=4, so -3=pi/36*1000*dx/dt, so dx/dt=-27/(250pi).