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Suppose the demand function of a commodity is q = 1000-5p, try to find the income function R (q) of the commodity, and find the income when the sales volume is 200 The answer is 32000. I don't know how to calculate it
P = 200-q / 5, so r (q) = Q * P = Q * (200-q / 5) = - Q ^ 2 / 5 + 200q when q is 200, r = 32000
Let the demand function of a commodity q = 1000-5p, then the income function R (q) of the commodity=
Revenue = sales * price
Because q = 1000-5p, P = (1000-q) / 5
Revenue function R (q) = P * q = (1000-5p) * (1000-q) / 5
=pq-1000p-200q+200000
=q*(1000-q)/5
For example, if the monthly sales q = 200 pieces is known, then R (200) = 32000
If the demand function is q = 5-4p, the demand elasticity function is (with problem-solving process)
From the question q = 80-4p, we can know that P = 20-0.25q. Because the income function should be the sales quantity multiplied by the unit price, the unit price is converted into the formula expressed in quantity, and then multiplied by the quantity, the expression of the unit price has been obtained, so r (q) = q (20-0.25q)
It is known that the total income function of an enterprise is r = 26q-2q ^ 2-4q ^ 3, and the total cost function is C = 8q + Q ^ 2 It is known that the total income function of an enterprise is r = 26q-2q ^ 2-4q ^ 3, and the total cost function is C = 8q + Q ^ 2, where Q represents the output of the product, Seek the output and maximum profit when the enterprise obtains the maximum profit
Answer:
The total revenue function is r = 26q-2q ^ 2-4q ^ 3, and the total cost function is C = 8q + Q ^ 2
Then the profit function:
y=R-C=26Q-2Q^2-4Q^3-8Q-Q^2=-4Q^3-3Q^2+18Q
Derivation of Q:
y'(Q)=-12Q^2-6Q+18
=-6*(2Q^2+Q-3)
=-6(2Q+3)(Q-1)
When - 3 / 20, y is an increasing function
When Q < - 3 / 2 or Q > 1, y '(q) < 0, y is the subtraction function
Therefore, when q = 1, the maximum y is y (1) = - 4-3 + 18 = 11
Therefore, when the output is 1, the maximum profit is 11
The marginal cost and marginal income function of a given product are C '(q) = q, respectively ²- 4q=6,R'(q)=105-2q It is known that the marginal cost and marginal income functions of a product are C '(q) = q respectively ²- 4q = 6, R '(q) = 105-2q, where q is the output, the fixed cost is 100, C (q) is the total cost, R (q) is the total income, and the output and maximum profit when calculating the maximum profit
MC = Mr, output when the maximum profit is solved
C = int MC DQ initial value condition C (0) = 100
R = int Mr DQ initial value condition R (0) = 0
Profit L = R-C
I think the title may have been copied wrong
Why is the minimum value of the function of subtracting the X-Power of 4 from the X-Power of 2 0?
Let a = 2 ^ x
Then a > 0
4^x=a ²
So y = a ²- a
=(a-1/2) ²- 1/4
a>0
So a = 1 / 2, the minimum value is - 1 / 4, not 0
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