A firm's product demand function is Q = 3P + 56.  What would be its total revenue function?
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
788734
3 ln p - ln q - 7 ln r\n= ln p^3 - ln q - ln r^7\n= ln ( p^3 /q) - ln r^7\n= ln(p^3 /(q r^7) ) \n\nx
403252
let x be the number from A\nlet y be the number from B\nlet z be the number from C\nthey have 500 pounds\nso\n2x+2y+3z=500\nand we have 1000 hours to produce \nso \n4x+5y+5z=1000\nsolve this system with respect to x u will get the following\n-2(2x+2y+3z=500)\n   4x+5y+5z=1000\nadd them u get \ny-z=0\nthen y=z\n5(2x+2y+3z=500)\n-2(4x+5y+5z=1000)\nu add and get\n2x+5z=500\nthen z=(500-2x)/5=100-2x/5\nz=y=100-2x/5\nlet f(x)=500x+600y+1000z=500x+y(1000+600) since y=z\n        =500x+1600y\nsubstitute y and z\nu will get\nf(x)=500x+1600(100-2x/5)=500x+160000-640x\n    =160000-140x\ndifferentiate f(x)\nu get\nf'(x)=-140\nso it is maximal when x=0\nthen y=z=100\nand profit =f(0)=160000$\nf(x)=
758175
Total money = $306,000\n\nRatio = 3:3:5:7  which means 1st = 3x ; 2nd = 3x ; 3rd = 5x ; 4th = 7x\n\nAdding 'x' would give you > 3x+3x+5x+7x = 18x\n\nEquation would be >  18x = 306000\n\nwhich will give  1x = 17000\n\n3x = 3*17000 (1st , 2nd)\n\n5x = 5*17000 (3rd)\n\n7x = 7*17000 (4th)\n\nThe rest should be pretty obvious.\n\n\n\nP.S. Leave a note if you need more help.
1138199
a) r = 4*MPK\n  w = 4*MPL\n\nb) Plug 2,1 into a) and rearrange:\nr:  (1/2)k^-1/2 l^1/4 = 1/2\nw: (1/4)k^1/2l^-3/4 = 1/4\n\ncancel out 1/2 and 1/4:\nr: k^-1/2 l^1/4 = 1\nw: k^1/2*l^-3/4 = 1\n\nfrom w:\nk^(-1/2)=l^(-3/4)\n\nstick this into r:\nl^(-3/4) l^1/4 = 1\nand solve\nl^(-1/2)=1\nl=1\n\nthen solve for k:\nk^(-1/2)=l^(-3/4)\nk^(-1/2)=1\nk=1\n\nso answer is that firm will use 1 unit of each input.\n\ncompute profit = 4*k^1/2l^1/4 - 2k -l\n=2
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~ ♥ ~
691620
(sigh)... Req = R1 + R2 + ...\n\nReq = 5 + 3 = 8.\n\nYay, I win.
509660
His profit will actually be found by taking his price minus all of his expenses - you can't just take the derivtive of the price function.  Since the price function given is per widget, multiply it by x to find the total revenue from the selling of x widgets.  The costs given are already given in terms of a total number of widgets, so we can just add those up.  So you're really looking at a profit function\n\nP(x) = x[90 - (x/24)] - (1200 + 60x + x^2/24)\n= 90x - (x^2/24) - 1200 - 60x - x^2/24\n= - 1200 + 30x - x^2/12\n\nTaking this derivative you get\nP'(x) = 30 - x/6\n\nTo find the max profit, set this to 0 and solve for x:\n0 = 30 - x/6\n30 = x/6\nx = 180\n\nSo after selling 180 widgets, you'll maximize profits.  And putting this back into the original per unit price function, you find that you're selling them at\n\nPrice = 90 - x/24\n= 90 - 180/24\n= $82.50 each
503348
No, the previous answer is wrong.\n\nGroup it like this:\n\n2p + pq -6q -3q^2\n\nthen factor like this:\np(2+q) -3q(2+q)\n\nfactor out the 2+q to get:\n(2+q)(p-3q)\n\nFOIL it out and you will see it is correct.
900057
a Q1 since it is between 0 and 67\nb Q2 since it is between 67 and 88\nc Q3 since it is between 88 and 105
1102694
intel_knight is almost correct on the supply and demand question.\n\nHere is what he said:\n\n> supply=demand\n> -5p+40 = -p^2+30p-8\n> p^2-35p+48 =0\n\nPerfect.  Now we need to apply the quadratic formula to do the problem.  a = 1, b = -35, c = 48, and the quadratic formula is\n\n(-b +/- sqrt(b^2 - 4ac))/2a\n\nSo it is (35 +/ sqrt(35^2 - 4*1*48))/2*1\n\nWhich will give two positive answers.  intel_knight, the mistake is that you have a -35 instead of a 35 at the beginning of the equation, -(-35) = 35.\n\np = (35 + sqrt(1225 - 192))/2 = $33.57\n\np = (35 - sqrt(1225 - 192))/2 = $1.43\n\nI would check the math on that for yourself, but that is what my calculator said.\n\nI suppose that they would want the higher number for greater profits.
53128
There is no mathematical formula.  It is what it is.  If you get 2% of all sales, then you get 2% of all sales.  Usually profit sharing only 1 or 2%.  Thats it.
285268
How are you at guessing?  You have to take in account the overall current market for the product or service you are trying to sell.  You have to take in account your ability to provide product.  You must take in account the minimum you must sell to support your business plan.  So pick a number that is reasonable to expect based on your products advantages over the competition, your ability to sell these advantages, your ability to deliver the product and whose sale numbers will support the business plan.
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
330608
As an accountant, you could take it a number of ways.\n\nfrom a basic trend analysis, 2, 3, ? = 4\n\nWhat about the sales push (advertising)? Or return customers? Word of mouth sales? \n\nThen what about new products? product obsolescence? \n\nExpansion efforts? New or better trained staff? New equipment available to increase production and therefore sales?\n\nThere are other factors beyond the simple data above that should be considered.\n\nCheers, L
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Let me try to give examples for when you use one, and when you use the \nother.\n\nThe profit of Company A, SYZO Ltd., has grown over the last three \nyears by 10 million, 12 million, and 14 million dollars. It is \nappropriate to say that it has grown by an average of 12 million \ndollars yearly, for which we use the arithmetic mean.\n\nThe profit of Company B, OZYS Ltd., has grown the over last three \nyears by 2.5%, 3%, and 3.5%. Here we cannot use the arithmetic mean \nand say that the average growth was 3%. Why not?\n\nSuppose that Company B, OZYS Ltd., started with a 100-million-dollar \nprofit. Three years later it will have become:\n\n     $100,000,000 * 1.025 * 1.03 * 1.035 = $109,270,125\n\nThis is less than a yearly increase of 3% would yield, since:\n\n     $100,000,000 * 1.03 * 1.03 * 1.03   = $109,272,700\n\nHere we see that we should use the geometric mean of the growth \nfactors 1.025, 1.03, and 1.035 to find the average percentage. That is \nalways less than the arithmetic mean would yield.\n\nI hope this helps.
841266
We're working on the domain [50,80].\n\nFor that domain, the rate per person is 200 - 2(x-50) = 300 - 2x\n\nYour total revenue is the rate per person times the number of people: (200 - 2x)x = 300x - 2x^2.\n\nYour profit is the revenue minus the total cost. Here, the cost has two components, a fixed cost and a variable cost.\n\nThe total cost is 6000 + 32x.\n\nTherefore, your profit function is (300x - 2x^2) - (6000 + 32x) =\n\nf(x) =  -6000 + 268x - 2x^2.\n\n To find the extrema of this function, take the first derivative and set it equal to zero. This isn't rigorous, but since f(x) is a parabola that points down, it will have a single extremum that must be a maximum. (You'd really have to use the second derivative test to verify this.) \n\nf'(x) = 268 - 4x = 0\n\n268 = 4x\n\nx = 67\n\nWith 67 people, the rate is $166. The revenue is $11122.\n\nThe costs are $6000 + 32*67 = $8144.\n\nThe profit is $2978.\n\nNote that the profit at x = 50 is $10000 - $7600 = $2400.\n\nThe profit at x = 80 is $11200 - $8560 = $2640.\n\nThis doesn't prove that we've found the maximum. But given what we know about f(x) - namely, that it's a parabola that points downward, we should expect that it's rising slowly enough for its vertex to be at an x-coordinate greater than x = 80, or that it peaks at a point between x = 50 and x = 80 that is closer x = 50 than it is to x = 80.\n\nJust as a second informal check, when x = 65, the revenue is $11050 and the costs are $8080, giving a profit of $2970.  Clearly, we're approaching the maximum that occurs at x = 67.