The cost of producing a number of items (x) is given by C = mx + b, in which b is the fixed cost and m is the variable cost (i.e., the cost of producing one more item.)\n\n(a) if the fixed cost is $40 and the variable cost is $10, what is the cost equation?\n(b) (in coordinates) how would graph the cost equation?\n(c) The revenue generated from the sale of x items is given by R = 50x. How would I graph the revenue equation on the same set of axes as the cost equation?\n\nand, finally,\n(d) How many items must be produced for the revenue to equal the cost (the break even point)?\n\nI HAVE tried to figure it out. I'm drained.\n\nThanks in advance....!
199904
(a) just substitute:  C = 10X + 40, where X = number of units produced.\n\n(b) It's a straight line with horizontal axis X and vertical axis C.  So, plot (0,40) and say, (5,90) and draw a straight line.\n\n(c) It's a straight line also thru the origin with a slope of 50.  Here, the vertical axis is R but it's the same scale as the C.  The line will go thru the points (0,0) and (1,50).\n\n(d) On the graph, it will be where the two lines intersect (1,50) or set the two equations equal to each other and solve:\n  \n     50R = 10X = 40  and 40R = 40  so X = 1
224715
This took me a while...\n\nSet up the equations like so:\nLet X be the cost per person\n\nXN=36000\n(x-1000)(n+3)=36000\n\nXN=36000\ntherefore\nX=36000/N\n\nSub that one into the other equation\n\n(36000/n-1000)(n+3)=36000\n(36000n/n)+(108000/n)-1000n-3000=36000\n36000+(108000/n)-1000n-3000=36000\n108000/n-1000n-3000=0\nmultiply both sides by N\n108000-1000n^2-3000n=0\nmultiply both sides by -1\n1000n^2+3000n-108000=0\n\nFACTOR THIS (it took me a while, i just tried a bunch of #s)\n\n(1000n+12000)(n-9)\n\ntherefore n=9,-12\n\nSince you can not have a negative # of people, the answer is n=9!\n\n:D\n\nAny questions email or message me
429753
It is my understanding that you would multiply the given costs by the surface area of those pieces to get total cost (C). Your equation would be as follows:\n2[(πr^2)(.027)] x πrh(.015)=C\nThe constraint equation is the volume:\n(πhr^2) = 16\nsolve your constraint equation for one of the variables.\nThen take the derivative of the cost equation and solve for your variable at "0." This will give you r (where r is radius) or h (where h is height of can) depending on what you solved for.
647298
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
383387
OK, first let's try to get that into a mathematical statement.  The volume can be given as yx^2, since the box is square, and the cost is 8x^2 for the top and bottom (4 cents times two sides), and 4xy for the sides (1 cent times four sides).  So the problem can be stated mathematically as yx^2 = 6912, c = 8x^2 + 4xy, minimize c.\n\nThe next step, I guess, would be to rearrange the equation to eliminate one variable and solve for c in terms of the other.  So say y = 6912/x^2, then c = 8x^2 + (4*6912)x/x^2.  That simplifies to c = 8x^2 + 27648x^-1.  Now we have c in terms of x.\n\nThe next step is to find the value of x for which c is minimized.  You don't say what math class you're in but the only way I know how to do this is with calculus, so I assume you are that advanced.  First you have to take the derivative of the function.  So the derivative of 8x^2 +27648x^-1 is 16x -27648x^-2.\n\nNext, find all points for which c' = 0.  Set 16x = 27648/x^2.  X = 1728/x^2.  x^3 = 1728.  x = 3_/1728 = 12.  So the only local min/max is at x = 12.\n\nNow, you would ordinarily compare the value of the function at the local min/max with the value at the boundaries, but since this function is undefined at the boundaries,we can assume that x = 12 is the answer.  \n\nGoing back to the first equation, yx^2 = 6912.  That means 144y = 6912.  y = 48.\n\nThe answer is (12, 48).
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Try this website for information on how to create a supply/demand graph:\n\nhttp://www.producingohio.org/lesson/draw.html
282945
Make this into a equation\nit takes $20 + $0.50 x (x being the number of treats)\nThe second equation is 1.50x to get the profit\nnow you set them equal to each other and solve for the intersection point
548386
If Profit = -4x^2 +24x -12\nthen we simply need to find the zeros of this function to determin the break even point(s)\n\n0=-4x^2 +24x -12\n0= -4(x-3)^2 +24\n-24/-4=(x-3)^2\nroot(6)=x-3\nx= 3+ root(6) OR 3-root(6)\n\n\nThe roots of this are at arround 0.5 and 5.5, so the firm will only make a profit if it sells between 1 and 5 units - the break even point would be one unit. \n\nIncidently this doesn't look like a very realistic profit formula!
668790
What is so great about this question. Answer is Rs. 1,142.86 Assume x to be selling price of each machine. so selling price of 10 machines is 10x. Since profit is equal to selling price of 3 machines (3x), therefore total selling price (10x) minus profit (3x) equals purchase price (Rs. 8000). Or in other words 10x - 3x=8000. Solve it to find out x (selling price) = Rs. 1142.86.
301559
This a calculus question?\n\nTo give you something to go on.....the question is poorly posed.\n\nYou sort of answer your own first question. You have a cubic equation, which, as I suspected, contains a max and a min (both considered 'local' from a calculus standpoint). The max is where your first answer comes from.\n\nThe next question is answered completely by saying that anything that is not the $40k will be the place at which the max starts to drop off. It doesn't ask for the MIN, it asks where it *starts* to drop off. This point is anything less that the $40k you stated in your question.\n\nHope this helps.\n\n:)
169304
The equation of a straight line (linear) is y =mx +c If you consider two variables (positions on a graph etc) y = m times X + a constant. On a graph m is the gradient measured as change in y over change in x and c is the constant (intercept) i.e the value of y when x =0\n\nso you are looking to produce an equation y = mx+c for your problem\n\nyou know that 200 items cost 22000-10000 so your m=12000/200\n\n=60. so if y=mx +c where y is 10000 and m is 60 and x is 100\n\nc= y/mx, c= 4000\n\nno use y =mx +c to check for x =300\n\n300*60 +4000 =2200\n\ny = 60x +4000
607369
Yes,\n\nThe equation relating p,s and c represents e very possible combination of T-shirts, caps and profit values for a funraiser. Suppose the class sets a profit goal of P=$600 . Finding cobinations of T-shirt and cap sales that meet this goal requires solving an equation with only two variables, s and c.\n\nA. Find five pairs of numbers for shirt and cap sales that will allow the students to make a $600 profit exactly.
381471
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
415951
Step 1: Find the Revenue per surgery, using only the amount of money spent on the surgery itself.\nPayment: $1000\nCosts: $380 in supplies, $15 in flowers, $20 * .25 in glasses ($20 * the 25% of people who need it = $400/patient\n\nRevenue = payment - cost = $1000 - $400 = $600/case\n\nStep 2: Find the annual costs.\nThe fixed costs are from the cost of the room itself (180K) and the salaries of the staff (270K).\nFixed costs: $180K + $270K = $450K.\n\nThere are no variable annual costs given in the problem, so you only need to worry about the fixed annual costs.\n\nStep 3: Find break even point.\nThe break even point is where the number of cases * the revenue/case equals the fixed costs.\nx * $600 = $450K\nx = $450K / $600 = 750 surgeries (solution)
525960
Hans spent 1.5x in dollars, where x is the number of programs bought.\n\nHans sold x - 20 programs, for a price of $3.00 each, which is a total of 3(x - 20) in dollars.\n\nThe profit of $15 is the difference between what he spent and what he sold, or 3(x - 20) - 1.5x.\n\nSetting all this up, 3(x - 20) - 1.5x = 15\n\nSolving for x, the number of programs he bought: \n\n3x - 60 -1.5x = 15\n1.5x = 75\nx = 50.\n\nYou can check your work:  50 programs cost Hans $75\nHans sold 30 for $3 each or $90.  $15 more.
594306
The two equations you are given are just two different ways to express the information.  The way this is set up assumes that the inventor will be able to sell every item that is produced, so both her production and sales are represented by "x".\n\nTo see that the two equations are really giving you the same information, notice that:\n\nSales - cost =profit.  Sales would be 4.00x.\n4.00x-(1.75x+7000)=2.25x-7000\n\nThe break even point is when your profit is zero, so you are neither losing nor gaining money.\nP(x)=0.  This point would be called the y-intercept if you were drawing a graph.\n\nAlso, if you look at your two equations, you can see that they are written in the form, y=mx+b, so you could graph the equation, and read the points off the graph, where:\n\n(a) x=2000\n(b) x=5000\n(c) the value of x when y=0.\n\nOr if you feel more comfortable with the algebra, just use your equations.\n\nThe best way is to find the answer by two different ways, and see if they agree.  For a quick check, you can just compare the answers and see if they make sense.\n\nJust by noticing that the slope is positive, you can already tell that the profit for 5000 units should be greater than for 2000 units.\n\nLittle ways to check your work are especially helpful during an exam.
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Revenue = Price x Quantity\nR = P*Q\n\nLet i be the increase in Price\nP = 40 + i\nQ = 5,000 - 100 * i\n\nR = (40 + i) * (5,000 - 100 * i)\n   = 200,000 + 5,000i - 4,000i -100i^2\n   = 200,000 + 1,000i - 100i^2\n\nDifferentiate R and set = 0 to find the maximum\n\ndR/di = 1,000 - 200i = 0\n 200i = 1,000\n      i = 5\n\nPlug this back in to the equation for Price = $40 + i\nP = 40 + 5 = $45\nQ = 5,000 - 100 *5 = 4,500\nR = P * Q = 45 * 4,500 = $202,500\n\nYou can calculate R at other prices ( try i = 4 and i = 6 ).  Then, you can verify that i = 5 is the MAX!\n\nThe important thing to learn is how to set-up the equations.  In this case we needed to create a new variable "i" in order to describe the price - quantity relationship.\n\nGood Luck\nm