A branch distributes a certain brand of products, and the cost of each product is 3 yuan, 10

1) The functional relationship between the profit l (10,000 yuan) and the selling price x of the branch for a year is:

l=(x-3-a)(12-x)^2,x∈[9,11]2)l'=0 to get x=6+2a 3 or x=12 (undesirable, discarded) 3 a 5, 8 6+2a 3 28 3

l on both sides of x=6+2a3'(x) only positive becomes negative.

So (1) when 8 6 + 2a 3 9 is 3 a 9 2, 2) when 9 6 + 2a 3 28 3 is 9 2 a 5, so q(a) = {9(6-a),3 a 9 2,4(3-1a 3) 3,9 2 a 5

So (1) when 8 6 + 2a 3 9 is 3 a 9 2, 2) when 9 6 + 2a 3 28 3 is 9 2 a 5,

1) l = (12-x)2 (x-3-a) 100002) above the formula, get.

l=[x3-(a+27)x2+(24a+216)x-144a-432]/10000

l derivative = [3x2-2(a+27)x+(24a+216)] 10000=

3x-2(a+9)](x-12)} 10000x-12<0 is constant, and when x<2(a+9), the function increments max=f[2(a+9) 3].

> ,..Subtract.

Let the function of l be f(x)).

Solution: (1) The functional relationship between the annual profit l (10,000 yuan) of the branch and the selling price x is:

l=（x-3-a）（12-x）2，x∈[9，11]．（2）l′（x）=（12-x）2-2（x-3-a）（12-x）=（12-x）（18+2a-3x）．

Let l (x)=0 get x=6+ a or x=12 (undesirable, rounded) 3 a 5, 8 6+ a

On both sides of x=6+ a, the value of l changes from positive to negative

So, when 8 6 + a 9, i.e. 3 a, lmax = l(9) = (9-3-a) (12-9) 2 = 9 (6-a);

When 9 6+ A, i.e., A5, lmax=l(6+ a)=(6+ a-3-a)[12-(6+ a)]2

4 (3-a) 3, that is, when 3 a, when the price of each piece is 9 yuan, the profit of the branch in a year l is the largest, and the maximum value q(a) = 9 (6-a) million yuan;

When A 5, when the selling price of each piece is (6 + A) yuan, the profit of the branch in a year l is the largest, and the maximum value of Q (A) = 4 (3- A) 30,000 yuan

1) According to the topic, first find the profit of each product, and then multiply it by the sales volume of one year, you can find the functional relationship between the profit of the branch for a year l (10,000 yuan) and the selling price x of each product;

2) According to the functional relationship between l and x, first find the derivative of the function, so that l (x)=0 can find the maximum profit when x=6 + 23a, and then according to the value range of a, the maximum profit is discussed when a takes different values

Solution: (1) The functional relationship between the annual profit l (10,000 yuan) of the branch and the selling price x is:

l=（x-3-a）（12-x）2，x∈[9，11]．

2）l′（x）=（12-x）2-2（x-3-a）（12-x）

12-x）（18+2a-3x）．

Let l (x)=0 get x=6 + 23a or x=12 (out of topic, discard).

3≤a≤5，∴8≤6+ 23a≤ 283．

On both sides of x=6 + 23a, the value of l changes from positive to negative

So, when 8 6 + 23a 9, i.e. 3 a 92, lmax = l(9) = (9-3-a) (12-9) 2 = 9 (6-a);

When 9 6 + 23a 283, i.e. 92 a 5, lmax = l(6 + 23a) = (6 + 23a-3-a)[12 - (6 + 23a)]2

4（3- 13a）3，q(a)={9(6-a)3≤a≤924(3-13a)392＜a≤5

That is, when 3 A 92, when the selling price of each piece is 9 yuan, the profit of the branch in a year l is the largest, and the maximum value is Q (A) = 9 (6-A) million yuan;

When 9 2 A 5, when the selling price of each piece is (6 + 3 2A) yuan, the annual profit of the branch l is the largest, and the maximum value Q (A) = 4 (3- 1 3A) is 30,000 yuan

I don't know you "sells (18-x) 20,000 pieces a year. (18-x)2 is not multiplied by 2......If yes! First question:

x-(3+a)] is the profit ...... of a productMultiplying the quantity (18-x) 2 gives us "l" i.e. ......l=:[x-(3+a)]*18-x)*2 Question 2: Simplify the above formula to get l=-2x 2+(42+2a)x-36a-108 and then use the vertex formula of the quadratic function to find the vertex ...... of the modified formulaAnd judge whether it is within the range of the value!!

The answer must be a ......And add the value range of a and x at the end!!

k Tai g i md 侑e Mo szhq intestine 4823644888k Tai g i md 侑e mo szhq intestine.

1) According to the topic, first find the profit of each product, and then multiply it by the sales volume of one year, you can find the functional relationship between the profit of the branch for a year l (10,000 yuan) and the selling price x of each product;

2) According to the functional relationship between l and x, first find the derivative of the function, so that l (x)=0 can find the maximum profit when x=6 + 23a, and then according to the value range of a, the maximum profit is discussed when a takes different values

Solution: (1) The functional relationship between the annual profit l (10,000 yuan) of the branch and the selling price x is:

l=（x-3-a）（12-x）2，x∈[9，11]．

2）l′（x）=（12-x）2-2（x-3-a）（12-x）

12-x）（18+2a-3x）．

Let l (x)=0 get x=6 + 23a or x=12 (out of topic, discard).

3≤a≤5，∴8≤6+ 23a≤ 283．

On both sides of x=6 + 23a, the value of l changes from positive to negative

So, when 8 6 + 23a 9, i.e. 3 a 92, lmax = l(9) = (9-3-a) (12-9) 2 = 9 (6-a);

When 9 6 + 23a 283, i.e. 92 a 5, lmax = l(6 + 23a) = (6 + 23a-3-a)[12 - (6 + 23a)]2

4（3- 13a）3，q(a)={9(6-a)3≤a≤924(3-13a)392＜a≤5

That is, when 3 A 92, when the selling price of each piece is 9 yuan, the profit of the branch in a year l is the largest, and the maximum value is Q (A) = 9 (6-A) million yuan;

When 9 2 A 5, when the selling price of each piece is (6 + 3 2A) yuan, the annual profit of the branch l is the largest, and the maximum value Q (A) = 4 (3- 1 3A) is 30,000 yuan

(1) Unit marginal contribution = 15-10 = 5 yuan.

2) Total marginal contribution = 5 * 15000 = 75000 yuan (3) Marginal contribution rate = 5 15 * 100% =

4) Variable Cost Ratio =

5) EBIT = 15 * 15000-10 * 15000-10000 = 65000 yuan.

Cost accounting? Let's go to a professional place and ask.

When investing x million yuan in advertising, the sales volume is 10y=x 2+7x+7 pieces, and the profit is 10y*(4-3)-x=x 2+7x-x+7=x 2+6x+7

When the profit is 160,000 yuan, x 2+6x-9=0 => x=[-6 (36+36)] 2=-3 3 2

x1=3 2-3 x2=-3 2-3 [Rounded] The advertising fee is 3 2-3, which is about 10,000 yuan.

1) Let the function be y=ax 2+bx+c

Data from the table.

c=1a+b+1=

4a+2b+1=

solution, a=-1 10, b=3 5, c=1

The function is y=-x 2 10+3x 5+1

2) According to the title, when the advertising fee is x (100,000 yuan), the sales volume will be y times the original sales volume, so the sales volume = 1 million * y

And because the cost is 2 yuan, the price is 3 yuan, and the profit Tongfan has to subtract the cost and advertising expenses.

And the original question bureau hail is 100,000 for 1 unit, so.

s=(3-2)*10*y-x=-x^2+5x+103)s=-x^2+5x+10

x-5/2)^2+65/4

When x=5 2, the function has a maximum value.

So (negative infinity, is the increasing interval of the function.)

Since 1 x 3, when 1 x is used, s increases as x increases

Solution: (1) Let the functional relationship between Lee's return belt y and x be y=ax2+bx+c, and the solution is: a=1a+b+c=, and the functional relationship between y and x is: y=;

2) Profit = total sales minus costs and advertising expenses, s = (3-2) 100y 10-x=-x2 + 5x+10;

3) s=-x2+5x+10=-(world roll, when x=, the function has a maximum value So x is the increasing interval of the function, since 1 x 3, so when 1 x, s increases with the increase of x

x = when the profit is the largest, the maximum profit is 100,000 yuan).

Let the cost price of each product be reduced by x yuan, according to the question, 510 (1 4%) 400 x) a(1 + 10%) = (510 400) m, and the solution is x =

The cost price of each product should be reduced to RMB;

The unit cost is the same as the unit selling price: 510 * 4% = yuan.

Suppose a company only produces and sells one product, and the unit price of this product is 20 yuan, and the normal production and sales volume each year.

BCD Variable Cost Sales Variable Cost Ratio, in the marginal contribution cost chart, if the independent variable is sales, then the slope of the variable cost line Variable cost ratio 12 20 100% 60%, therefore, option A is incorrect. In the state of capital protection (i.e., under the critical point of profit and loss), profit is 0, sales volume fixed cost (unit price unit variable cost) 2400 (20 12) 300 (pieces), and the utilization degree of the enterprise's production and operation capacity 300 400 100% 75%, therefore, the statement of option B is correct. The margin of safety 1 75% 25% falls within the range of 20% 30%, so option d is correct.

Marginal contribution of clever states in the margin of safety Margin of safety Marginal contribution rate (400 20 300 20) (1 60%) 800 (yuan), so option c is correct.