Calculate a formula for price-elasiticy of demand, and simplify for D(p)=600/p^3.?
969075
price elasticity of demand = (dD/dP) * (P/Q)\n\nwhere D = the quantity demanded, P = price, and Q is the quantity.\nP/Q is the price to quantity ratio at whatever point you are trying to calculate elasticity.\n\nSo then price elasticity of demand = (-1800/P^4) * (P/Q)  \n\nwhen p = 25, Q = 600/(25^3) = 0.0384\n\nso price elasticity of demand when P = 25 is:\n(-1800/25^4) * (25/0.0384) = -3
626617
5. D\n\n($6. * 40 hrs) + ($6. * 1.5 * 6.5 hrs) + ($6 * 2 * 8 hrs)\n=$394.50\n\n\n6. B\n\n13 * $250. = $3250. Value sold @ 5% = $162.50 commission\n 3 * $250. = $750.  Value sold @  3% = $22.50 commission\nTotal Commission = $185.00
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
757602
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
475512
a) P(D+|K+) = P(D+ and K+)/P(K+)\n17/30 = P(D+ and K+)/.6\nP(D+ and K+) = .34\n\nb) P(D- | K+) = P(D- and K+)/P(K+)\n=P(D- and K+)/.6\n=(P(K+) - P(D+ and K+))/.6\n=.26/.6\n= 13/30
715515
Profit- P(x)\n\nP(x) = R(x) - C(x)\nP(X)= 20x - .001x^2 - 2500 - 10x\nP(x)= 10x - .001x^2 - 2500\n\n\nWe have our equation, let's differentiate it (i'll use the shortcut and the power rule to speed things up).\n\nP'(x) = 10 - .002x\n\nThe zeroes of this will be local extrema of the original equation.  Since there is a negative sign in front of the .001x^2, that flips the parabola upside down and it will be the maximum point.\n\n0 = 10-.002x\nx=5000\n\nPlug this into the cost equation:  2500 + 10(5000) = 2500 + 50000 = 52500.\n\nHope this helps.
1160809
1000 ppm/1,000,000 x 100% = 0.100%\n\nIts a simple division solution.
687032
cost\nsaturated market
161170
2p,10q\nis this ur homework?\n*****
33091
3m(-6m-2m^2 p - 10p) = 3m(-6m) - (3m)(2m^2 p)- 3m(10p)\n= -18m^2- 6m^2 p - 30mp
871291
This is a great estimate for this problem!  Good job!\n\nIf you want an exact answer, you can just set it up a little further.  You already have that the weekly salary is $150 + 6% of sales.\n\nIf we let 'x' be the amount in sales and convert 6% to .06, we can derive the equation\n\n150 + .06 x = 650,\n\nwhich is the equation to find out how much in sales the employee needs to make exactly 650.\n\nThen we do algebra:\n\n.06 x = 500\n\nx = 500/(.06) = 8333.333333. . .\n\nSo your estimate was really close.  Keep up the good work!
1130953
Take the original/wholesale price, and divide it by .35\nThis will equal your selling price. (with 65% markup)
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.
513457
Law of Supply and Demand:\n\nIn a nutshell - If more people want something the more expensive it will be until there is more produced. When there is an excess amount produced (more than people want) Then the price will be less.
358099
0.43 & 0.0062\n\nDivide by 100
1049343
Let x = product 1, y = product 2, and z = product 3\nP = 50x + 20y + 25z\n9x + 3y + 5z <= 500\n5x + 4y + 0z <= 350\n3x + 0y + 2z <= 150\n"SALES POTENTIAL FOR PRODUCT 1,2 EXEEDS THE MAX PRODUCTION AND FOR3. 20PERWEEK"\nI'm not quite understanding this sentence.\nWhat is the max production?\n\n"AND FOR3. 20PERWEEK" I interpret that as z >= 20
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.