Maximize the objective function subject to the given constraints.?
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is this for a algebra class? calculs class? matrix methods?\n\nTry this.\n\nWe want to simplify the complicated list of 3 constraints to two simple constraints: a range for y and a range for x.\n\nThe lower bound is simple: constraint III says all x and y are greater than or equal to zero.\n\nFind a range for y. Algebra can be done on these eqations the same as with an equal sign as long as the alligator is in the correct direction (it is). So eliminate x from the equations by using, "Constraint II minus Constraint I"; or;\n\n  x+3y<_6\n-(x+ y<_2)\n\n=2y<_4....... or, y<_2. Therefore, 0<_y<_2.\n\nSimliarly for range x, eliminate y by "Three times Constraint I plus Constraint II.", or:\n\n3*(x-y<_2) =  3x- 3y<_6\n                    +( x+3y<_6)\n                    = 4x<_12.... or, x<_3. Therefore, 0<_x<_3.\n\nNow we simply have an equation of three variables with two constraints.\n\nz=3x+2y;  0<_y<_2\n                 0<_x<_3\n\nI'm not sure what you want from here, but I think I understand.\n\nThe graph will be a plane in the x,y,z coordinate system.\n\nThe maximum z is achieved by using maximum values of x and y.\n\nThe corners will be defined by the extreme values of x and y with the correseponding z.\n\nFor instance, with (x,y,z)...\n\n(0,2,z1)\n(3,0,z2)\n(0,0,z3)\n(3,2,z4)\n\nThis will create a "polygon...?" in the x,y,z plane with the corners at the locations shown above.\n\nThis might be able to be done more simply by using matrices but I can't see an obvious way.
605315
Occular strength X Objective strength.
1137202
Cosidering that the load (2P) has been applied to the midpoint of the bar, the load which will be carried by each cantilever beam will be 'P'.\nIn that case, the maximum deflection at the free edge of the system will be  \n\n(P*L*L*L )/ (3EI)\nwhere L is the span of each beam \nEI is the flexural rigidity
142854
Linear programming is explained in various ways in this site..from there you can link to some examples also..\n\n\nhttp://www2.isye.gatech.edu/~spyros/LP/node1.html#SECTION00010000000000000000
690146
Optimisation problems are normally done using partial derivatives, however you're problem is discreet so it will be handled differently. First make a map of the network lets say with coloured lines to represent the individual paths from the beginning to the end of the manufacturing process and nodes where they intersect, i.e. nodes that represent each machine. Then there are multiple ways to run the algorithm. For the case of minimum number search through the nodes to see if there is one that bundles all the lines through it if not start looking for pairs that would have all the lines going through one or the other of the nodes and so on until you find a solution. Unless visually it is a simple graph you will need a computer to do the processing.\nThe problem with a cost assigned is more difficult. In this case each node has a value associated with it. In this case generate an ordering from cheapest to dearest of the possible costs of selecting any number of nodes. The test each of these in succession until you find one that corresponds to having all the lines pass through at least one of the selected nodes associated with that particular cost.\nIn knowing details of the way in which the graph would look or the cost for certain locations it may be possible to modify the algorithm so it runs quicker, but this should do it for the information given.
739585
egalisation! 2 partout!
577894
Plyometrics with weights.
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) = ...
773761
I believe the answer is biconvex and elliptical.
650534
Maximum public benefit at minimum public cost.
826268
2l + 2w = 300\n2l = 300 - 2w\nl = 150 - w\n\nMaximize length times width:\n\nA(w) = w(150 - w) = 150w - w²\n\nTo maximize, take derivative, set = 0, and solve for w.\n\n150 - 2w = 0\n2w = 150\nw = 75\n\nand then l = 150 - 75 = 75.
832985
Let S = x + 1/x.\n\nWe want to minimize x for x>0.\n\nTake the derivative of S with respect to x and set that equal to zero to find the extrema of S:\n\n0 = dS/dx = 1 - 1/x^2\n\n1/x^2 = 1\n\n1 = x^2\nx = 1 (and -1, but we are only interested in x>0 here).\n\nWe can check to make sure this is a minimum by looking at the sign of the second derivative of S with respect to x at x=1:\n\nd2S/dx2 = 2/x^3\n\nWhen x = 1, the second derivative is equal to 2, which is positive, so the extremum at this point is a minimum (the function is concave upward).
749284
You need to take partial derivatives of your cost function (w.r.t. x and w.r.t. y) and solve the resulting equations to get critical points. Then, generally, you need to check whether your function is maximized or minimized at the critical points..\n\nIf partial derivatives are not covered in the class, try to argue that x = y (a square achieves the maximum area of all rectangles of a given perimeter).  With this, you have only one variable in the cost equation; use ordinary differentiation...
399858
Correct, derive and solve for x=0, that will give you the maxima or minima.\nUsing product rule:\ndy/dx= 2ln(x)*(1/x).\nSolve for 2ln(x)/x=0; *x both side\n2ln(x)=0; /2 both sides\nln(x)=0; \nThe answer is 1. The natural log of 1 is zero.\nx=1\n\nTherefore, the min or max is at 1. \n\nIf you want to know if its a max or min you take the second derivative.\n\nd^2y/dx^2= 2/x^2-2ln(x)/x^2=2-0=2 -> If evaluated at the min/max value of x=1.\n\nSince the second derivative is positive, this function is convex, and at 1 it is minimized.
1077864
Assuming D is z=24x+8y\nThe max with those numbers would be C with a Z value of 24.5 times 25 or 612.5
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Try starting with a proportion. Let K be the cost we are looking for.\n\nb / x = K / y\n\nNow solve for K.\n\nK = by / x
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omg its a calculus 1 optimization problem.