find where the revenue function is maximized using the given linear demand function?
440148
x=30\n\nRevenue is max when xp is max.\n\nxp=Rev=-0.2x^2 + 12x\n\nThe derivative is given by -0.4x+12 which = 0 when x = 30\n\nRev at x=30 is 180
152511
Which function, Subsitution, Linear COmbination, Graph?
84748
production function Q = 6L -0.5L squared\nI'm basing this on Q = 6L - (0.5L squared)\nand not Q = (6L - 0.5L) squared\n\nI would have to assume the owner wants to have the maximum production, therefore that Q should be the highest number.\n\nWhen L = 1, Q = 5.75.  \nL = 2, Q = 11.  \nL = 3, Q = 15.76\nL = 4, Q = 20\nL = 5, Q = 23.75\nL = 6, Q = 27\nL = 7, Q = 29.75\nL = 8, Q = 32\nL = 9, Q = 33.75\nL = 10, Q = 35\nL = 11, Q = 35.75\nL = 12, Q = 36\nL = 13, Q = 35.75\nSo the correct answer is L (employees) = 12
919994
Find the 2nd derivative and find the max.
1026879
f(x,y)= (1-x)^2 + 105 (y- x^2)^2\n\nIt's often used to test out various nonlinear optimization algorithms.
889254
You must graph the inequalities and see where the vertices of the resulting polygon are.  Then plug the coordinates of them into your storage space function and see which gives the highest amount.\n\nLet x = # of cabinet A and y = # of cabinet B\n\nx>= 0\ny >= 0\n3x + 6y <= 60 (square feet inequality)\n75x + 50y <= 600 (budget inequality)\n\nStorage function: S = 12x + 18y
648089
http://en.wikipedia.org/wiki/Law_of_supply_and_demand
705660
Revenue = Price x Quantity: R = P * Q\nDefine C = Cost \nDefine Q = Quantity which you call X in your equations.\n\nCost is defined by the equations:\nC = 360,000 - 90P\nC = 90,000 + 30Q\n\nput these equations together and \n\n360,000 - 90P = 90,000 + 30Q\n\nSolve for Q\n\n270,000 - 90P = 30Q\n9,000 - 3P = Q\n\nNow, go back to the original Revenue Equation:\nR = P * Q\n\nR = P * (9,000 - 3P)\nR = 9,000P - P^2  a quadratic function of price P\n\nNote:  if P = 0, R = 0.  Then R increases until P = 4,500.  This is the maximum R which you can find by differentiating R, setting = 0 and solving for P.  After this point, if we keep increasing P,  R coninues to fall back to 0 at P = 9,000.\n\nNote:  there is a little problem, here.  Your original equations state:\nC = 90,000 + 30X and X = 9,000 - 30P  Substitute for X\nC = 90,000 + 30 * (9,000 - 30P)\nC = 90,000 + 270,000 - 900P\nC = 360,000 - 900P\n\nTHIS IS DIFFERENT FROM YOUR EQUATION: C=360,000-90P\nOne equation says 900P, the other says 90P.\n\nThe math is the same.  Depending on which is correct, your answer will change!\nGood Luck!
352638
You're missing part of the question here.\n\nFor the demand function given, at x = 100:\nd(x) = 200 - (100 * .5)\nd(x) = 200 - (50)\nd(x) = 150\n\nBut you haven't given enough information to calculate consumer surplus.
896473
Here's the method (do the calculations yourself).\n\nP(x) the price at which you can sell the quantity x\nR(x) = xP(x)  [we don't know the price yet, it depends on the unknown x]\nProfit = R(x) - C(x) = xP(x) - C(x)\n\nTo maximize the profit over x, work out its derivative with respect to x, equate it (the der.) to 0, and solve for x. That's the optimum (profit-maximizing) quantity.\n\nIf you also want to know the price and the maximum profit, put the x you've found into the formula for P(x) = ... . Finally, put this x into the formula for the profit: it is xP(x) - C(x) = ...
690202
production, distribution, and consumption of goods and services
861163
First you graph the feasible region.  Let x be the number of burritos sold, y the number of nachos.  10<= x <= 40.  This gives you a vertical band, shading between a vertical line crossing the x-axis at 10 and one crossing the x-axis at 40.  Likewise 30<= y <= 70 gives you a shaded band between horizontal lines that cross the y-axis at 30 and 70.  Where they overlap you have a rectangle.  The final constraint is x+y <= 90, which slices off the top right corner of the rectangle.\n\nNow you need to know the coordinates of the corners.  Bottom left is (10,30), top left is (10,70), bottom right is (40,30), upper top right is (20,70) and lower top right is (40,50).\n\nFinally, your profit function is P = 0.95x + 0.45y.  It's slope is a little steeper than -2, and will reach a maximum at a corner of the feasible region far from (0,0).  \n\nTry it at the upper top right: .95(20)+.45(70) = 50.50\nTry it at the lower top right: .95(40)+.45(50) = 60.50.  So max profit with 40 burritos and 50 nachos.
512935
(a)\nC = mx + b\nC = 10x + 40\n(where fixed cost could be due to the cost of the workers or rental of the factory...)\n(The variable cost is the cost required to make 1 item)\n\n(b)\nR = 50x\n(The revenue, money collected, from selling x items is 50x.\nSo if you were to buy 1 item, you would pay $50\nand if you were to buy 3 items, you would pay $150.)\n\nRevenue = Cost\nR = C\n50x = 10x + 40\n50x - 10x = 40\n40x = 40\nx = 40/40\nx = 1\n\nThe selling of 1 item would result in the break-even point.
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) = ...
519282
fast moving consumer goods
923838
Do a quick back-of-envelop calculation as follow\n\nMonthly expenditure (rent + car+ food + utilities + travel) * 12 * 25
251685
First things first:\nyou MUST  know what are you going to solve for. Do you want to maximise return given a threshold for variance? Do you want to maximise financial results or mark-to-market results?\n\nIn either case, do you have a set of future value expectations? Do you have a covariance-matrix.\nYou should do the following:\n1) Specify (model) your problem\n2) See what data you can count on. If necessary, go to step \n3) research what methods are there available for solving your problem\n4) ONCE YOU GET THERE research software that allows your desired method, complexity and size.\nSmall problems, no matter how "complex" they are, can be solved by Excel.\nHowever, even simple problem, if large enough, need specialised software.