CuZ Tuppl Iunctin ‘gwven by nduhi Qud MI AMurttb wnujominb> Pra -Pomnt ldlul cattainr FTudalct Ctco 1760 0.6! and total [etenuctA ~gienby R(x) enFtT GuJDM Rreak-cuen points If total costs fof _ EuknmAlin arc given by Clx)- Z3rand tolal are given hrlcn QUILIC> Marunizing revenue a7dl profit Flnd the maximum Icvenuc And MXIUMT Fiont for the functions described Problcm Ereul h und prufit manmhizanor Glcn Foju 0.0lx 30, find maximum profit and the break-even quantities and sketch the graph profit Given C(x) 0.21 aud R(x) 0.217, find the level of production that gives maximum profit and find the maximum Profit Kteui-ert HHIA profit maximization etnannecomie pany has fixed costs of SLS,000 for its product and vari” 0.04x dollars per unit There able costs given by ris the total number of units The selling price of the product / given by 3J00 0.06x dollars Per unit Formulate the functions for total cost and total Fcaenc T0%” Itknumtrt Rast I What type of function is this? the function Product Ricent 0 US Aulu #uh dodo In tht Hu 10zot explains its [X25) cAnin” Ehelaahuidacata ol pbxrosynthesis that enalnCLN depchtOn tinnt ICcnnany PDc [ Iorz % Graph this function with graphing – Eculaltt (UreA-MLO Cnnang -T7-1 And only when J(} Ithal ! Tlr mudel 1buit IFr Xa) ne I-values MenaQcat Suppose the weekly cost C, in dollars eliminates Percent of the Pollution from the hntelancon} given 4800? Khzl N ol [unction this? Glcnatip Tcntetents LIC Right of Pollution Removed Wal D the domain of Clp)? Find C(0) and inlet pret meaning Find the weekly removing 9933 of the pollution Municipal water costs The Municipal Water of Bella Isla used the following function chme I #lci 10.x 7*s0] Clx) = 1023 8,16(x 100) 100 < x 8367 6.76x ICCO) if > > 1000 where Clx) the cost dollars hou water Find the monthly charge for 12000 gl Find the monthly charge for 825,00 water:
The Correct Answer and Explanation is:
1. Total Cost and Revenue Functions
You mentioned a company with fixed costs and variable costs per unit. The general form for Total Cost (C(x)) and Total Revenue (R(x)) is:
- C(x) = Fixed costs + Variable cost per unit * x
- R(x) = Price per unit * x
Given:
- Fixed costs = $15,000
- Variable cost = $0.04 * x per unit
- Price per unit = $30 – 0.06 * x (which is a linear decrease in price as more units are sold, this could be a typical case in price discrimination or bulk selling)
Thus, the functions become:
- C(x) = 15,000 + 0.04x
- R(x) = (30 – 0.06x) * x = 30x – 0.06x²
2. Maximizing Profit
Profit is defined as Profit(x) = Revenue(x) – Cost(x).
To find the maximum profit, we need to:
- Express the Profit function: P(x)=R(x)−C(x)=(30x−0.06×2)−(15,000+0.04x)=29.96x−0.06×2−15,000P(x) = R(x) – C(x) = (30x – 0.06x^2) – (15,000 + 0.04x) = 29.96x – 0.06x^2 – 15,000P(x)=R(x)−C(x)=(30x−0.06×2)−(15,000+0.04x)=29.96x−0.06×2−15,000
- Find the critical points by taking the derivative of the profit function and setting it equal to zero: P′(x)=29.96−0.12xP'(x) = 29.96 – 0.12xP′(x)=29.96−0.12x Setting P′(x)=0P'(x) = 0P′(x)=0 to find the value of xxx: 0=29.96−0.12x⇒x=29.960.12≈249.670 = 29.96 – 0.12x \quad \Rightarrow \quad x = \frac{29.96}{0.12} \approx 249.670=29.96−0.12x⇒x=0.1229.96≈249.67 This means the optimal production level is approximately 250 units.
- Find the maximum profit:
Plug x=250x = 250x=250 into the profit function: P(250)=29.96(250)−0.06(2502)−15,000=7,490−3,750−15,000=−11,260P(250) = 29.96(250) – 0.06(250^2) – 15,000 = 7,490 – 3,750 – 15,000 = -11,260P(250)=29.96(250)−0.06(2502)−15,000=7,490−3,750−15,000=−11,260 This suggests that the company is incurring a loss at this production level. You would need to check other cost structures or revise the pricing model to maximize profits.
3. Break-Even Points
Break-even points occur when profit equals zero, i.e., R(x)=C(x)R(x) = C(x)R(x)=C(x). To find this: 30x−0.06×2=15,000+0.04x30x – 0.06x^2 = 15,000 + 0.04x30x−0.06×2=15,000+0.04x
Rearranging terms: −0.06×2+30x−0.04x−15,000=0-0.06x^2 + 30x – 0.04x – 15,000 = 0−0.06×2+30x−0.04x−15,000=0
Simplifying: −0.06×2+29.96x−15,000=0-0.06x^2 + 29.96x – 15,000 = 0−0.06×2+29.96x−15,000=0
Solve this quadratic equation using the quadratic formula: x=−29.96±(29.96)2−4(−0.06)(−15,000)2(−0.06)x = \frac{-29.96 \pm \sqrt{(29.96)^2 – 4(-0.06)(-15,000)}}{2(-0.06)}x=2(−0.06)−29.96±(29.96)2−4(−0.06)(−15,000)
After calculating, you’ll find the break-even quantities (the production levels at which total revenue equals total costs).
4. Pollution Removal Costs Function
You also mentioned a pollution removal scenario where the cost function is given. For example:
- C(x) = 1023 * (x – 100) when x > 1000
This function describes the cost of removing pollution based on the number of units xxx removed. You could calculate costs for specific values of xxx and sketch the graph accordingly.
5. Monthly Charges for Water
For the water charges, assuming a tiered pricing model:
- C(x) = 1023 + 8.16(x – 100) for x>1000x > 1000x>1000, where xxx represents the number of gallons used.
- If you need to calculate the monthly charge for 12,000 gallons or 8,000 gallons, simply substitute those values of xxx into the cost function and solve.
For instance:
- For 12,000 gallons: C(12,000)=1023+8.16(12,000−100)=1023+8.16(11,900)=1023+97,344=98,367C(12,000) = 1023 + 8.16(12,000 – 100) = 1023 + 8.16(11,900) = 1023 + 97,344 = 98,367C(12,000)=1023+8.16(12,000−100)=1023+8.16(11,900)=1023+97,344=98,367
- For 8,000 gallons, repeat the same process with x=8,000x = 8,000x=8,000.
