how  to figure thebox containing the greatest volume, the sum of girth   the length equals 108ins.?
598778
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
972145
1) You need to solve these equations first...\n\nW*L = 192\nL*H = 128\nH*W = 96\n\nLet's multiply the first two equations by each other:\nW*L * L*H = 192 * 128\nH*W * L^2 = 24576\n\nNow divide this by the last equation:\nL^2 = 24576 / 96\nL^2 = 256\nL = 16.\n\nNow you can figure the rest of the dimensions:\nW*L = 192\nW*16 = 192\nW = 192/16\nW = 12\n\nW*H = 96\n(12)*H = 96\nH = 96/12\nH = 8\n\nSo the dimensions are:\n8 inches high, 12 inches wide, 16 inches long.\n\nThat's 2 cubes high (2 x 4" = 8"), 3 cubes wide (3 x 4" = 12"), 4 cubes long (4 x 4" = 16").\n\nAll together that is 24 cubes (2 x 3 x 4 = 24)\n\n2) If you shrink everything by a factor of 2.54 (to get centimeters) the answer doesn't change at all.  Since every dimension changes to cm, the box shrinks and so do the cubes.\n\nYour dimensions will be 8 cm x 12 cm x 16 cm.  With a block that is 4 cm on each side, you still fit 2 blocks x 3 blocks x 4 blocks = 24 blocks.
493858
this will be  169646.00329384883487698274269709 cubic inches
755525
OK, first let's try to get that into a mathematical statement.  The volume can be given as yx^2, since the box is square, and the cost is 8x^2 for the top and bottom (4 cents times two sides), and 4xy for the sides (1 cent times four sides).  So the problem can be stated mathematically as yx^2 = 6912, c = 8x^2 + 4xy, minimize c.\n\nThe next step, I guess, would be to rearrange the equation to eliminate one variable and solve for c in terms of the other.  So say y = 6912/x^2, then c = 8x^2 + (4*6912)x/x^2.  That simplifies to c = 8x^2 + 27648x^-1.  Now we have c in terms of x.\n\nThe next step is to find the value of x for which c is minimized.  You don't say what math class you're in but the only way I know how to do this is with calculus, so I assume you are that advanced.  First you have to take the derivative of the function.  So the derivative of 8x^2 +27648x^-1 is 16x -27648x^-2.\n\nNext, find all points for which c' = 0.  Set 16x = 27648/x^2.  X = 1728/x^2.  x^3 = 1728.  x = 3_/1728 = 12.  So the only local min/max is at x = 12.\n\nNow, you would ordinarily compare the value of the function at the local min/max with the value at the boundaries, but since this function is undefined at the boundaries,we can assume that x = 12 is the answer.  \n\nGoing back to the first equation, yx^2 = 6912.  That means 144y = 6912.  y = 48.\n\nThe answer is (12, 48).
635866
I might be missing something, but the cubic space something occupies is also referred to as it volume, where:\n\nfor a six sides regular box shape volume = heightxbreadthx depth\n\n( make sure all your measurements are in feet eg 1 foot 5 inches is 1 and 5/12 feet. \n\nHowever this formula is not true for a sphere or a trapezoidal box or a pyramid.  For whcih I'd recommend wikipedia or the old maths textbook as its hmmm, err 18 years since I last did all that stuff.
732533
D = M/V\n\nDensity= Mass/Volume
883381
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
796040
material needed=2pir(h+r)\nvolume=pir^2h\npir^2h=1000cc (given)\nh=1000/(pir^2)\nsubstituing h in the equation 2pir(h+r)\nmaterial=2pir(1000/pir^2+r)\n=2pir*1000/pir^2+2pir^2\ndA/dt=-2000/r^2+4pir\nsetting this to zero\n-2000+4pir^3=0\n4pir^3=2000\nr^3=2000/4pi\nr=cube root of 500/3.14\n=5.4 approx\nsubstituting\nh=1000/(3.14*5.4*5.4)\n=10.9 aapprox\nso the radius is 5.4 cm approx\nand the height 10.9 approx
783613
Area = length x width
706050
Ok... the first thing to realize is that you are finding the maximum area with the least amount of materials.  To do that you should put the shed at a corner of the rectange, that way two sides of the shed act as part of the fence.  Assume that the shed is not "usable area" for the shep pen.  Accordingly, you need to solve the problem twice, once with using the shed as part of one side, and once using the shed as a corner.\n\nOk, the first thing you need to do is show that the maximum area of a rectangle occurs when the lenght and width are equal.  That is the only part that needs any type of equation analysis, and it really doesn't require calculus.\n\nA=L*W\nAssuming that the width is at most as long as the L, then the equation goes to:\nA=L*(x*L), where x a percentage, ranging from 0 to 1.\nClearly, the area (A) will be maximum when x=1.  That shows that in your problem, the pen will have the maximum area when the length and width of the rectangles are equal.\n\nSo, you know that the fence is 160m., and you have 4 sides, s1, s2, s3 and s4.\nSo:\n160m=s1+s2+s3+s4\n\nNow, if you use the shed as part of one side, namely s4, then the perimeter equation for the shed is:\n180m=s1+s2+s3+s4+20m\n\nSince the maximum area occurs when the lenght and width are equal:\ns1=s2=s3=s4-20m, so substituting s2, s3 and s4 for s1 in the permiter equation, you get:\n4s1+20m=180m\nSolve for s1,\ns1=45m (call this the length)\ns4=45-20=25m\nin this case, the area will be S1*S1, since the lenght and width of the pen will be equal.  \nSo the area is 45*45=2025 sq. m.\n\nNow, put the shed in a corner of the sheep pen.\nSo in our example, we'll make the shed part of s1 and s4.\n\nso the primeter equation will be:\ns1+20m+s2+s3+s4+20m=200m\nNow, s1=s2-20, s3=s2 and s4=s2-20\nSo substituting for s1, s2 and s4, you get:\ns2-20m+s2+s3+s4-20m+40m=200m\nsolve for s2,\ns2=50m=s3 and s1=30m (s2-20m)\nSo max area is 50m*50m=2500 sq. m\nBut, now you need to reduce the area by the area of the shed (400 sq. m), so final usable area is:\n2100 sq. m which is more than using the shed as part of only 1 side.
486266
the formula for volume of a cylinder is V = h * pi * r^2.\nAlternatively, the surface area is A = h * pi * r * 2.\n\nSo, if you have either the surface area or the volume, you can calculate for the height.\n\nH = Volume / pi * r^2\nH = Area / pi * r * 2
67968
It is similar to calculating the area. \n\nIt is a product of Length and Width (L x W).
880761
People above neglected the vertical component of 460 N.  This must be added to the weight of the box.  Also, not sure why they took 350 N and multiplied it by acceleration.\n\nso F_vert=460sin35.6=267.78 N (downward)\n     F_hor=460cos35.6=374.0 N\n\ncalculating F_frict=mu x (F_normal)\n\nF_normal=weight of box + F_vert = 267.78+350=617.78\n\nso F_frict=0.57x617.78=352.13 N\n\nnow the net force on the box = 374.0 - 352.13 = 21.87 N\n\nand we need mass of the box, m=350/9.81=35.7 kg\n\nso a=21.87/35.7=0.6126 m/s2\n\nfinally,  d=V_initial x t + 0.5 x a x t^2, V_initial=0\n\nso t=((4.2)/(0.5x0.6126))^0.5=3.7 sec\n\n**********************************************************\n\nHowever, should you wish to ignore the vert component\n\nF_frict=mu x (F_normal)=0.57 x 350 = 199.5 N\n\nso net horizontal force is:\n\nF= 374.0 - 199.5 = 174.5 N, a=174.5/35.7= 4.89 m/s2\n\nplug a into:\n\nd=V_initial x t + 0.5 x a x t^2, and solve for t\n\nt=((4.2)/(0.5x4.89))^0.5 = 1.31 sec
192399
In the front?  You have to consider the whole cube to find volume.  Volume = base area * height  You could find the surface area of one side, but you'd need to know the side length.
618423
A cylinder is round, therefore the largest diameter would be 7. The depth of the box is 2, so the height of the cylinder would be 2. the volume of a cylinder is found by multiplying the height of the cylinder by the area of its top (pi r squared). Now get back to your homework.
253154
l = length\nw = width\n\nl = 8w\n\nlw = 128\n(8w)(w) = 128\n8w² = 128\nw² = 16\nw = 4\nl = 8(4)\nl = 32\n\nWidth is 4 feet.
760290
Popcorn ball. . . . \n\nGoogle search:  volume sphere (should have used "volume sphere," but it turned out not to matter)\n\nhttp://www.aaamath.com/geo79-volume-sphere.html\n\n4/3 (pi) r^3\n\n4/3 times pi times (6 to the 3rd power) \n\nequals 4/3 times pi times 216. . . and work from there.\n\nGoogle search:  volume cylinder\n\nhttp://www.321know.com/exp79_x5.htm\n\n(area of the base) * height =(pi r squared times height)\n\nPi times (16 squared) times 275. . . work from there.