Optimization problem?
411206
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.
26357
You don't need linear programming for this...\nYou want to maximize the investment in C, then in B, then in A.\n\nHe should in invest $6,000 in bond C.\nThen he can invest only $2,000 in bond B.\nThen he can invest only $8,000 in bond A.\n\nHe'll have $2,000 remaining.\n\n\nEdit to add:\nSorry, my LP skills are a bit rusty.  Here's all I remember:\nMaximize .05*A + .75*B + 1.05*C;\nsubject to   \nA + B + C <=18000\nA + B <=10000\nB + C <=8000\nC <=6000
258345
The efficiency of an ideal engine is (1- T2/T1) or (T1- T2)/T1 = (600 - 400) / 600 =0.333.\nBut in the problem it is given that it is 0.5 which is not possible.
783604
volume=pir^2h=1000\nsince there are two variables we need a minimum of two equations
836789
All this problems are solved using the equation:\n\nHeat = M * Cp * DeltaT\n\nPlug the numbers, check wich one is OK.  Oh, and watch the units.
573900
hey dude i hav d same problem blive me n still havnt figured out d solution for dis so can any1 one help me......\nif u get d solution den email me at arein1986@yahoo.co.in
900562
I'd suggest searching for such problems here on yahoo answers.  You can find out how to do it.
977689
Can you post the actual problem here, so we can give you specific help?
774817
I can tell that nobody wants to just do this problem.  There's nothing interesting about it.  It's a typical problem that a stat student does to show understanding.  Get out your textbook and study up how to solve it.
960933
Revenue= price times quantity\nRevenue= x(-4x+28)=-4x^2+28x\nProfit = Revenue - Cost\nProfit= -4x^2+28x-4x-12 = -4x^2+24x-12\nIn order to find the quantity that maximizes profit, you should take the derivative of the profit function and equate it to zero.\nDerivative of Profit= -8x+24=0 So x=24/8=3
783536
let the length and width be 'x' and 'y' reply.\ngiven xy=512\nso x=512/y\nthe perimeter=2x+y\n=2x+512/x\nfor optimising\ndP/dx=0\n2-512/x^2=0\nmultiplying by x^2\n2x^2-512=0\n2x^2=512\ndividing by 2\nx^2=256\nx=rt256=16\ny=512/x=512/16=32\nso the dimensions for min perimeter\n=16'*32'\nperimeter=2x+y=64'\ncost=64*2=$128
691515
algorithms, data structures, computer organization
993077
not that I know...you might have to wright your own algorithm!\n\nouch...have fun!\n\nbut check links!!
982689
Here's the method (you can do the calculations yourself, surely).\nMaximize the profit xp(x) - C(x) over x: work out d/dx of the profit, set it equal to 0, and work out x. That's the optimum (profit-maximizing) quantity. Then put this x into the formula for p(x) = ... to work out the price. Finally, put this x into the formula for the profit: it is xp(x) - C(x) = ...
1066492
x = -4\n10points thank u!
364028
Richard Bellman made this principle in regards to a two-stage optimization to get V(t+s, x, φ)\n\n   Using any U up to a time t, with a controller U changes to an optimal one.   At time t+s we get a performance J(t,x,V(s,∙,φ),U).  If you take the supremum with respect to U you get V(t+s, x, φ)
208332
What problem are you having? You did not provide any code, or even a possible algorithm. Try actually providing us with some better information, so that we can actually help you.