Demand function p = 10-Q / 5, cost function C = 50 + 2q, output, total profit is the largest

Demand function p = 10-Q / 5, cost function C = 50 + 2q, output, total profit is the largest

When the output is Q, P = 10-Q / 5, then the sales revenue p * q = (10-Q / 5) * Q
The cost is C = 50 + 2q
Maximum profit p * q-c maximum q = 20

Let the total cost function and the total revenue function be
1、 Let C = C (q) = (2 / 3) Q ^ 3-10q ^ 2 + 36q + 3 be the total cost function and the total revenue function respectively
R=R(Q)=4Q
Seek the output level, product price and profit at the time of maximum profit
2、 The total cost function of a factory producing 1000 pieces of a certain product every day is C = (1 / 2) Q ^ 2 + 36q + 9800 yuan,
In order to minimize the average cost, what is the average cost of each product when the daily output is several?
Good answers will be awarded another 50 points

1. Let y (q) = R (q) - C (q) = - 2 / 3Q ^ 3 + 10q ^ 2-32q-3,
Y'=-2Q^2+20Q-32,
Let y '= 0, we get Q1 = 2, Q2 = 8,
From Q > = 0, so [0,2], (8, + infinity) is the decreasing interval and [2,8] is the increasing interval. When q = 8, ymax = 39.67,
Output level 8, price 4, profit 39.67
2, from Q > = 0,
Average cost per piece = C / Q = 1 / 2q + 36 + 9800 / Q > = 36 + 2 radical (1 / 2q * 9800 / Q) = 36 + 140 = 176,
If and only if 1 / 2q = 9800 / Q, i.e. q = 140, the lowest cost per piece is 176

Seeking quick return
The price of a certain brand bottled beverage is 26 yuan per case. A store promotes the bottled beverage with "buy one get three free (buy one get three free)". If you buy the whole case, buy one get three free, which is equivalent to 10 / 13 of the original price. How many bottles are there in a case of this brand beverage?

There are 10 bottles
Let the original price be x yuan,
26/x=26/(10x/13)-3
The calculated x = 2.6
So there are 26 / 2.6 = 10 bottles

Find a mathematical problem!
Party A and Party B go to a store to buy the same commodity twice, the price is different each time, and the way of purchase is different. Party A buys 1000 kg each time, and Party B buys 1000 yuan each time
(1) What is the average unit price of the goods purchased by Party A and Party B?
(2) Which is the more cost-effective way of purchasing?

1) Suppose the unit price of the first commodity is m yuan / kg, and that of the second is n yuan / kg,
A bought a total of 2000 kg, using 1000 m + 1000 n yuan,
So the average price of a: (1000m + 1000N) / 2000 = (M + n) / 2
B bought a total of 1000 / M + 1000 / N kg, with 2000 yuan
So the average price of B: (1000 + 1000) / [1000 / M + 1000 / N] = 2Mn / (M + n)
2) (M + n) / 2-2mn / (M + n)
=(m-n)^2/2(m+n)
According to the meaning of the title, m and N are not equal,
So (m-n) ^ 2 > 0
So (m-n) & # / 2 (M + n) > 0
So the second way is more cost-effective

There are two bookshelves. 14 books in bookshelf A is equivalent to 25 books in bookshelf B, and there are 120 more books in bookshelf a than in bookshelf B______ Ben

25 / 14 = 85, 85-1 = 35120 / 35 = 200 (copies); answer: there are 200 books in B bookshelf

A math problem, quick return
Four pieces of wood are combined to form a parallelogram with a base of 24 cm and a height of 16 cm. If you pull it into a rectangle, the area is 72 square cm larger than that of the parallelogram, how many cm is the perimeter of the parallelogram
Remember to bring units. Oh, no units,

Area of parallelogram: bottom * height = 24 × 16 = 384,
So the area of the rectangle is 384 + 72 = 456, and the length of the bottom is 24cm,
Then the width of the rectangle is 456 / 24 = 19,
So the perimeter of the parallelogram is 2 (24 + 19) = 86 cm,

Ask a math problem
A certain Bomber can fly at a speed of 1540 km / h in the downwind and 1260 km / h in the upwind. When it is full of fuel, it can fly continuously for 10 hours. After each mission, it must leave enough time to return to the base, otherwise it will crash on the way. When it starts in the downwind, it can fly continuously for 10 hours, What is the farthest attack distance? Suppose the wind direction does not change during the mission

If x hours of downwind flight is set, 10-x hours of upwind flight is set
1540 x = 1260 x (10-x) the distance along the wind is equal to that against the wind
1540X=12600-1260X
280X=12600
X=4.5
Flight distance = 1540x4.5 = 6930

Seek all trigonometric identity transformation formula, including various variants, universal formula!

Basic relations of trigonometric functions with the same angle
Reciprocal relationship:
Business relationship:
Square relation:
tanα ·cotα＝1
sinα ·cscα＝1
cosα ·secα＝1
sinα/cosα＝tanα＝secα/cscα
cosα/sinα＝cotα＝cscα/secα
sin2α＋cos2α＝1
1＋tan2α＝sec2α
1＋cot2α＝csc2α
Induction formula
sin（－α）＝－sinα
cos（－α）＝cosα tan（－α）＝－tanα
cot（－α）＝－cotα
sin（π/2－α）＝cosα
cos（π/2－α）＝sinα
tan（π/2－α）＝cotα
cot（π/2－α）＝tanα
sin（π/2＋α）＝cosα
cos（π/2＋α）＝－sinα
tan（π/2＋α）＝－cotα
cot（π/2＋α）＝－tanα
sin（π－α）＝sinα
cos（π－α）＝－cosα
tan（π－α）＝－tanα
cot（π－α）＝－cotα
sin（π＋α）＝－sinα
cos（π＋α）＝－cosα
tan（π＋α）＝tanα
cot（π＋α）＝cotα
sin（3π/2－α）＝－cosα
cos（3π/2－α）＝－sinα
tan（3π/2－α）＝cotα
cot（3π/2－α）＝tanα
sin（3π/2＋α）＝－cosα
cos（3π/2＋α）＝sinα
tan（3π/2＋α）＝－cotα
cot（3π/2＋α）＝－tanα
sin（2π－α）＝－sinα
cos（2π－α）＝cosα
tan（2π－α）＝－tanα
cot（2π－α）＝－cotα
sin（2kπ＋α）＝sinα
cos（2kπ＋α）＝cosα
tan（2kπ＋α）＝tanα
cot（2kπ＋α）＝cotα
(where k ∈ z)
Trigonometric function formula of sum and difference of two angles
Universal formula
sin（α＋β）＝sinαcosβ＋cosαsinβ
sin（α－β）＝sinαcosβ－cosαsinβ
cos（α＋β）＝cosαcosβ－sinαsinβ
cos（α－β）＝cosαcosβ＋sinαsinβ
tanα＋tanβ
tan（α＋β）＝——————
1－tanα ·tanβ
tanα－tanβ
tan（α－β）＝——————
1＋tanα ·tanβ
2tan(α/2)
sinα＝——————
1＋tan2(α/2)
1－tan2(α/2)
cosα＝——————
1＋tan2(α/2)
2tan(α/2)
tanα＝——————
1－tan2(α/2)
Sine, cosine and tangent formulas of half angle
The power reduction formula of trigonometric function
Sine, cosine and tangent formula of double angle
Sine, cosine and tangent formula of triple angle
sin2α＝2sinαcosα
cos2α＝cos2α－sin2α＝2cos2α－1＝1－2sin2α
2tanα
tan2α＝—————
1－tan2α
sin3α＝3sinα－4sin3α
cos3α＝4cos3α－3cosα
3tanα－tan3α
tan3α＝——————
1－3tan2α
Sum difference product formula of trigonometric function
Sum difference formula of trigonometric function
α＋β α－β
sinα＋sinβ＝2sin—－－·cos—－—
2 2
α＋β α－β
sinα－sinβ＝2cos—－－·sin—－—
2 2
α＋β α－β
cosα＋cosβ＝2cos—－－·cos—－—
2 2
α＋β α－β
cosα－cosβ＝－2sin—－－·sin—－—
2 2 1
sinα ·cosβ＝-[sin（α＋β）＋sin（α－β）]
two
one
cosα ·sinβ＝-[sin（α＋β）－sin（α－β）]
two
one
cosα ·cosβ＝-[cos（α＋β）＋cos（α－β）]
two
one
sinα ·sinβ＝－ -[cos（α＋β）－cos（α－β）]
two
The form of trigonometric function which transforms asin α ± bcos α into an angle (the formula of trigonometric function of auxiliary angle)

Derivation of universal formula for trigonometric identity transformation

What is the universal formula?

sinα=[2tan(α/2)]/{1+[tan(α/2)]^2}
cosα=[1-tan(α/2)^2]/{1+[tan(α/2)]^2}
tanα=[2tan(α/2)]/{1-[tan(α/2)]^2}
The substitution of sin α, cos α and Tan α into Tan (α / 2) is called universal substitution