Please help me set up this optimization problem?
831548
V = 2hw², h = 48/w²\nAb = 2w²\nAs = 2wh\nAe = wh\nAl = 2(2wh + wh) = 6wh = 6*48/w\nC = 12*2w² + 5*6*48/w\ndC/dw = 48w - 1,440/w² = 0\n48w = 1,440/w²\nw³ = 30\nw = 3.1072 m\nh = 48/3.1072² \nh = 4.9716 m\nC = $781.44
1142827
You don't give a feasible region. However, only one of your answers actually satisfies z=6x+10y. That should solve your problem.
319998
http://www.math.iitb.ac.in/resources/manuals/algorithms.html click d link to know. It may be helpful to u.
351221
Have you ever heard about ANT COLONY OPTIMIZATION?\nAnts are much more wiser than we think...\nHave a look at this: \nhttp://www.aco-metaheuristic.org\nyou'll enjoy it.
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) = ...
691515
algorithms, data structures, computer organization
17380
What you want is called "Search Engine Optimization". There are several do-it-yourself sites out there, check out the Source below.
751869
First, we need to find the partial derivatives of cost with respect to length and width.\n\nC = 10LW + 640(1/W + 1/L)\ndC/dL = 10W - 640 * (1/L^2)\ndC/dW = 10L - 640 * (1/W^2)\n\nThe minimum of cost will occur where the above partial derivatives are equal to 0.\n\n10W - 640 * (1/L^2) = 0\n10W = 640 * (1/L^2)\nW(L^2) = 64\n\n10L - 640 * (1/W^2) = 0\n10L = 640 * (1/W^2)\nL(W^2) = 64\n\nSo we have 2 equations which both need to be true at minimum cost\n\nW(L^2) = 64 and L(W^2) = 64\nL = 64/(WL) and W = 64/(WL)\n\nSo W and L must be the same\n\nW(W^2) = W^3 = 64\nW = L = 4 meters\nH = 2 cubic meters / (4 * 4 square meters) = 0.125 meters\n\nNote that the given equation is correct, Gary needs to substitute\nhw = 2/l and hl = 2/w to get the final result...
368963
Hello i dnt think its possible but u can try with AIEEE this is equilent to B.E
1154622
a variable quantity that can be resolved into components
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
1180637
I was answering another similar question, and I came across a website that has a downloadable sample spreadsheet for doing statistical distributions.  Play with this and see if it helps you out.
181915
set up variables for everything.\n\nFor example: 7200 cars.  3200 four seaters.  then the rest is 7200-3200=4000 six seaters.  the 4 seaters are 100% full so 40000 people.  the total is 80% full thats 80%*(4000*6 + 3200*4) etc
725388
let them be x and y\n\nxy = 588\nwe need to minimize x+3y\nlet 3y be z\n3xy = xz = 588*3 = 1764\n\nminimize x+z\n(x+z)^2- 4xz = (x-z)^2\n(x+y) is minimum when (x-z)^2 is minimum or x = z\n\nso x=z = 42\nso x= 42 y = 14
763428
QUESTION1 -- Profits = Revenue - Cost, so you want to come up with an equation for profits using R(x) and C(x), then use differentiation to determine your maximum.  Let P(x) = profits.  So:\n\nP(x) = R(x) - C(x) =\n50x - 0.5x² - 9x - 3 =\n-0.5x² + 41x - 3\n\nP '(x) = -x + 41\nP '(41) = 0  << so you have a min or max at x=41\n\nP ''(x) = -1\nP ''(41) = -1  << tells us x=41 is a max\n\nSo, answer for #1 = Maximum profit at x=41\n\nQUESTION 2 -- Let x = side of base, so L and W each = x since the base is a square.  The height is unknown, but we can solve for it since we know the volume.\n\nV = 57 ft³ = x²h,\nso h = 57/x²\n\nWe want to maximize surface area, so let's first get our formula (all in terms of x) for the surface area:\n\nS(x) = x² + 4(x*57/x²) = x² + 228/x\n\nMaximize same way we did above using differentiation:\n\nS '(x) = 2x -228/x²\n\nS '(x) = 0 when 2x = 228/x², i.e., when x³ = 114; i.e., when x = cube.rt(114) ft.\n\nSo the dimensions of your box are L = W = x = cu.rt.(114), and h =  57/x² = 57/114^(2/3) ft.\n\n~ ♥ ~
1037424
evenly distributed energy
899962
Let x equal the cost of each ball.........\n\n5x </= to 20\n\nNow try your hand at solving it.