How many units do 1 Kw solar panels produce in 5 hours of sun?
Go through the calculations below:-1kW = 1000 wattsIf the availabilty of sun is of 5 hoursThen, 1000/5 = 200 watts1 Solar panel is of 200 watts.Then no of solar panels = 1000 / 200 = 5Then 5 solar panels of 200 watts is being used.And generally it will produce 4–5 units per day.I hope you got your answer.One of the leading pioneers in Solar and renewable resources known as ZunRoof, a brand which you can trust on. A team of IIT alumni started ZunRoof a couple of years ago, but it has created its own name in a very short period of time. Go Solar with ZunRoof.We take our advice for installing solar power systems and reach us out for any kind of queries.You can reach us at ZunRoof
How many units should be produced during the month? accounting help pls!?
1. parker company's budget indicates that 18,000 units are to be sold during september. if the company has 3,000 units on hand at the beginning of the period and plan to have 4,000 units on hand at the end of the period at the end of the month, how many units should be produced during the month? Beginning Inventory + Units Produce - Ending Inventory = Units Sold 3,000 + Units Produced - 4,000 = 18,000 Units Produced = c. 19,000 2. a suit company has the following standards to make one suit: standard quantity standard price direct materials 4 yards per unit $9.50 per yard direct labor 2 hours per unit $12.00 per hour the company used 13,000 yards of material in order to make 3,000 suits in april. the direct materials quantity variance is: Standard 4 x $9.50 x 3,000 = $114,000 Actual (13,000 / 3,000) x $9.50 x 3,000 = $123,500 123,500 - 114,000 = b. $9,500 unfavorable 3. a suit company has the following standards to make one suit: (note this is a different question from #2) standard quantity standard price direct materials 4 yards per unit $9.50 per yard direct labor 2 hours per unit $12.00 per hour the company purchased 4,000 yards of material in march for $40,000. the company used 3,800 yards in march in order to make 900 suits. the direct materials quantity variance is: Standard 4 x $9.50 x 900 = $34,200 Actual (3,800 / 900) x $9.50 x 900 = $36,100 36,100 - 34,200 = b. $1,900 unfavorable
How many units should be produced to maximize profit?
R'(x) = 200 -10x R(x) = Integral of R'(x) = 200x -10x^2 /2 +c1 = 200x-5x^2 +c1 C'(x) =40-5x C(x) = Integral of C'(x) = 40x-5x^2 /2 + c2 Profit = P(x) =R(x)-C(x) = ( 200x-5x^2 +c1) -( 40x-5x^2 /2 + c2) P(x) = 200x -5x^2 +c1 -40x +5x^2 /2 -c2 To maximize the profit, differentiate the profit with respect to x P'(x) = 200 -10x-40 +5x = 0 -5x =-160 x=32 units should be produced P''(x) =-10+5 =-5 < 0, so profit has been maximized.
How many units per year is produced from 1mw solar plant?
The generation of Solar plant depends on many factors:1) Location 2) Solar irradiation3) Climatic ConditionsNormally the generation will be 1500 kWh(5 hours X 300 days). Normally we take 300 days per year as solar days. Depending on the location the hours of operation per day will vary.For South Indian plains the generation from 1 MW plant will be 1000 kW X 5 hours X 300 days = 15,00,000 kWh
Determine how many (thousands of) units must be produced to yield maximum profit. Determine the maximum profit?
a) P(x) = -x² + 40x - 18 You can determine the maximum profit by taking the derivative of P(x), setting it equal to zero, and solving for "x": P'(x) = -2x + 40 = 0 -2x = -40 x = -40/-2 = 20 (thousands of units) Solve for P(20): P(20) = -(20)² + 40(20) - 18 = -400 + 800 - 18 = 382 (thousands of dollars) b) P(x) = -x² + 40x - 18 = 40 -x² + 40x - 58 = 0 The roots of the equation are found from the quadratic formula where, from the above equation, a = -1, b = 40, and c = -58: x = {-b ± √[b² - 4ac]}/2a = {-40 ± √[(40)² - 4(-1)(-58)]}/2(-1) = 20 ± (1/2)√(1,368) = 20 ± 3√38 x = 1.507, 38.493 Thus, you must produce more than 1.507 thousand units and less than 38.493 thousand units.
How many units are generated by 1 kW solar panel?
A 1kW solar panel would generate 4.0 to 4.5 units in a normal day. However, as the power generated by a solar system is dependent upon the solar radiation which the panels receive, the units generated may vary at any given day. For example, the units generated may be less, if the geographical location doesn’t experience bright sunshine.The generation may also be lower, if the panels blocked by shadow objects or if the climate/weather is not suitable, soiling and etc. Since solar is driven by technology and requires maintenance, the energy output is also largely dependent on how well the solar system is engineered.This is why a specialized solar partner such as MYSUN is so important for consistent performance. To know the recommended system size for the best solar experience, run the MYSUN solar calculator and calculate your savings.
A manufacturer can sell a certain product for $80 per unit. Total cost consists of a fixed overhead of $4,500 plus production cost of $50 per unit. How many units must the manufacturer sell to realize a profit of $900?
A manufacturer can sell a certain product for $80 per unit. Total cost consists of a fixed overhead of $4,500 plus production cost of $50 per unit. How many units must the manufacturer sell to realize a profit of $900?This seems quite straightforward. He makes $30 on each item out of which has to come the overhead and the profit that total $5400. Divide this by $30. I get 180. What do you get?
A manufacturer has daily production costs of C=.55x^2−110x+10,000 where C is the total cost (in dollars) and x is the number of units produced. Determine how many units should be produced each day to yield a minimum cost. What is that minimum cost?
$4500. Taking the derivative of C(x) yields C'(x) = 1.1x-110. Setting this equal to zero to find the min/max value of x yields x=100. Plugging 100 back into C(x) yields C(100) = $4500. C''(x) = 1.1 which is positive for all values of x, so that this implies that that C(100) is a minimum value.