A model for a company's Revunue is R=-30p^2+420p+18,000 , where p is the price in dollars of the company'd product. What price will maximize this revenue? Find the maximum revenue.
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take the derivative and solve for 0 and you get \n\n-60p+420=0... \n\nthat works out to p=7.  plug it into your original formula and you get \n\nR=-30(7^2)+420(7)+18,000=19470\n\nand $19470 is your max revenue, with 7 being revenue maximizing price
436014
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.
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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.
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Let X be the sales Kenisha need to make, the equation will be \n     0.04X + 500 = 0.05X = 400\n     500 - 400 = 0.05X - 0.04X\n     100 = 0.01X\n     10000 = X\nI think the first offer is better, cause it kind of tough to have $10000 sales per month, you think? Hope this help.
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P= x* (price per unit) - x*(costs per unit) - fixed costs\nP > 0\nx*(price per unit - costs per unit)> fixed costs\nx> fixed costs/ (price per unit - costs per unit)
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(a) If marginal revenue is greater then the marginal cost, this signifies that variable costs for production of every subsequent unit are greater then additional revenue associated with it. Therefore, the company is either operating above the equilibrium point or is unable to reach it. As such, company should reduce the level of production to achieve the equilibrium point or stop producing at all. \n\nI would love to help you more on this assignment. If you need professional help - contact us at http://custom-writing.org/\nYour source for professionally written essays and A+ homework.
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Let us assume that we rent out x videos.\nTherefore, for the first video shop, the cost is 15+2.5x.\nFor the second video shop, the cost is 5+3x.\n\nLet us equate them so that the cost becomes equal.\n15+2.5x=5+3x.\n.5x=10\nx=20\nTherefore, you need to rent more than 20 videos from the first shop to make it more economical.
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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
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Let final vol. = v\n\nSo, P(v-30) = 885\n\nNow just substitute P in N/m^2 and 30cm3 = 30*10^-6 m3
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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) = ...
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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!
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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)=
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1. The demand gives you the price of a single item, so the revenue will be the price of an item times the number of items bought:\n\nr(x) = x * p(x) =  -4x^2 + 28x\n\n2. The profit function is revenue - cost:\n\n(-4x^2 + 28x) - (4x + 12) = -4x^2 + 24x - 12\n\n3. First, it would be helpful to factor out a common factor:\n\n-4(x^2 - 6x + 3)\n\nIf you have a quadratic function x^2 + bx + c, in order for it to be a perfect square, c must equal (b / 2)^2. In this case, b = -6, so c should equal 9. So, adding 6 (while also subtracting 6 to not change the value of the expression) should give a perfect square.\n\n-4(x^2 -6x + (3 + 6) - 6)\n\n-4( (x^2 - 6x + 9)  - 6 )\n\n-4(x^2 - 6x + 9) + 24\n\n-4(x - 3)^2 + 24\n\nThis puts the equation into vertex form, so the vertex or maximum value occurs where x = 3.
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Cost to produce/market the product.\nGoing rate for similar products.
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Basically this is a class on "what if" scenarios.  "What if the company raises the price of a widget by $1.25?  What will that do to our sales?  What is the projected increase or decrease in sales stated in terms of a percentage?"\n\nIt is more complicate than that but that is essentially it.
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ok, there is not a lot of information to help you but it sound like you want a general framework.   If a company wants to grow revenue, they have two options:\n\n1) Grow existing base of revenue:  The key question here is how do you grow sales to existing customers.   What are the obstacles to additional sales?   You should develop ideas for how you remove these barriers.   Can you change the product/service so it is more appealing?   Can you change pricing to drive additional revenues?  Can you increase the price in a way so it does not impact your unit volume?  Can you decrease price so it increases the unit volume of sales?   Can you change your service or sales channels to faciliate increased frequency of sales?\n\n2) Develop new sources of revenue:   You can either sell new products to your existing customers or obtain new customers.  You should develop new ideas around these two areas.   Based on the products/services you sell, what other products is your set of customers looking for?   Would they buy those products from you?   Can you sell accessories or value added services?   You can also look for new customers. Which customer types drive the most revenue?  How do you find more of them?   Can you increase your sales force or your marketing budget in order to reach them?
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One could buy eight horses and chariot sets and then sell them as suggested in the multiple choice to see how it works or one could set up two sets of equations and solve:\n\n-10%*(Horse Price) + 20%*(Chariot Price) = 0 (no profit)\n 10%*(Horse Price) + -5%*(Chariot Price) = 15 (profit)\n\nCombine the equations to get:\n                     -15%*(Chariot Price) = 15\n\nSolve for Chariot Price... 100\n\nPlug back into either equation to get Horse Price... 200\n\nThe answer is c