for demand function d(x)=200-0.5x find consumers surplus at the demand level x=100?
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.
200344
W=f.delta(x) \nx=w/f = 49821/717= 69.49 metes
549479
The prior answer is exactly right, but as far as a general rule goes:\n\nDivide the current value by the present value - here, $200,000 divided by $100,000 = 2\n\nIf you want to figure out the average increase over x periods, simply take the xth root of the quotient you calculated above.\n\nHere, again, for example, you're looking for the 5th root of 2, because you're trying to figure out the average increase over 5 years (therefore, the 5th root) of 2 (the quotient)\n\nSo - using that formula here:\n\nThe 5th root of 2 (the same as 2 raised to the 0.2 power) equals 1.14869.\n\nIgnore the leading 1 - you're looking for the numbers after the decimal place. Therefore, the average rate of growth is 14.869 percent per year over 5 years.\n\nThis approach should work for any similar problem.\n\nGood luck!
519289
let the original price be x\nx * 20/100 =12590\ntherefore.. x=12590*5\n=\n62950\n\ni didnt notice the extra f in "off" .. i took it as "of".. \nstill m not goin to edit my answer
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.
762202
The manufacturer's total profit is given by his revenues minus his costs\n\nFor 0 < x <= 3 tons, the revenue is $13/ton * x while his costs are x^2 + 5x + 7.  His net total profits are therefore:\n\nP[0<x<=3] = (13x - x^2 - 5x - 7) dollars\nP[0<x<=3] = (-7 + 8x - x^2) dollars\n\nThe maximum of the profit function can be found by taking the derivative of P with respect to x setting it equal to zero, and solving for x:\n\n0 = dP/dx = 8-2x\n\nx = 4\n\nThis maximum is outside the range for which this profit function is valid, so the actual maximum profit for x<=3 tons occurs right at x=3 tons.\n\nFor x > 3 tons, we have that the profit is given by:\n\nP[x>3] = (13x - x^2 - 5x - 7 - 3(x-3)) dollars\nP[x>3] = (2 + 5x - x^2) dollars.\n\nNote that at x=3, both profit functions have the same value.\n\nLet's again find the maximum of this profit function:\n\n0 = dP/dx = 5 - 2x\n\nx = 2.5\n\nIn this case, the maximum of this function occurs at a value  of x below the minimum value for which this function is valid.  This implies that the maximum accessible value of this profit function occurs as x -> 3.\n\nThe manufacturer's profit is maximized when he makes 3 tons of the item.  \n\nThe easiest way to check that this is the correct answer is to plot the two profit functions over their valid ranges and see that the maximum does, indeed, occur at x=3.
489532
Let x=Box-office tickets No.\nthen since pre-purchased tickets No.=180\nthen your equation is:\n4.5*180+8*x=6*(x+180) solving by x results x=135
382763
Divide 950 by 5\nAnswer: $190
79486
demand for resource depends on the price of that resource... either the actual selling price (if the price was too high, consumers would be reluctant to buying it) or the opportunity costs in obtaining the resource (the difficulty in obtaining the resource.. or what you must give up in order to get that resource)\n\nim currently taking a microeconomics class too.
701603
2. 200 \n(75/100)x=150\nx=200\n\n3. 10 %\nx =(75/750) x 100\nx=10%\n\n4. 5000\n(6.5/100)x=325\nx=5000\n\n5. 724.5\n(48.3/100) x 1500\n\nHope this helped!!!
903752
first, find the slope of the equation\n\nslope = change in y-value/change in x-value\n\nslope = (24,500-11,000)/(500-200) = 45\n\nthen, find the equation-- do this by plugging in the values for x anf y to find c, the constant\n\ny = 45x + c\n\n11,000 = (45 x 200) + c\n\n11,000 = 9,000 + c\n\nc = 2,000\n\nNow, you have your equation:\n\ny = 45x + 2,000 \n\nNow, it's easy, just plug in 800 for x. This gives you:\n\ny = (45 x 800) + 2000\n\ny = 38,000\n\nIf this is not worth 10 points, I don't know what is...\n\nBoth of the people above me are wrong. I am sure of it, I am a math major in university. Even though my answer may seem a little strange, because it means you pay $2,000 for no books, trust me, it is correct.\n\nI know $2000 is the setup fee, I just couldn't think of the right word.
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
246875
You haven't really given equations, but maybe you mean to say that the equations are \nQd=50-(price) Qs=1/2(price)-10\n\nConsumer surplus is the area of the triangle with points of the form (q,p) at (10,40), (0,40) and (0,50). The area of this triangle is $50.\n\nProducer's Surplus is the area of the triangle with points at (10,40), (0,40) and (0,20). The area of this triangle is $100. \n\nCheck your equations again and we can give you a better picture.\n\nGood luck!
462280
E = Fd = (2E-9 N) (200E-12 m) = x J\nx J (1 kJ / 1000 J) = xE-3 kJ\n\nI think this is already per mol...
32878
200\n\n(1,000,000/5,000=200)
809580
The first part is an increase of 100%\n\nYou started by using 10 units, so 10units represents 100%.\n\nnow you add another 10 units so you use 100 + 100 % of original amount\n\nso you use 200% of original amount.  You started by using 100% so you subtract 100% from 200% to get 100% increase\n\nThis example following makes the working clearer.\n\nyou start using 10 units.\n\nThe fist quantity in thes sums always represents 100%\n\nSuppose you increase your consumption to 30 units\n\nThe increase in consumtion = 30 - 10 units \n\n=20 units\n\nthe orignal consumption = 10 units reps 100%\n\nso increase, 20 units = 100 x 20/10%\n\n=200%
86569
The total revenue function is equal to the number of units sold (in this case, P) times demand function, Q.\n\nThe revenue function R(P) is equal to P time Q(P).\n\nR(P)= P(3P + 56)\nR(P) = 3P^2 + 56P