SOLUTION: A firm's fixed cost is $50,000, the cost of making 10,000 units of its product is $100,000 and the cost of making 20,000 units is $250,000. Find the firm's cost function if it is k

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Question 1042307: A firm's fixed cost is $50,000, the cost of making 10,000 units of its product is $100,000 and the cost of making 20,000 units is $250,000. Find the firm's cost function if it is known to be quadratic.
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
it took some doing, but i think i found the answer.
the answer is y = .0005 * x^2 + 50,000

the graph of that equation looks like this:

$$$

the procedure is as follows:

the general form of a quadratic equation is:

y = ax^2 + bx + c

when y = 100,000 and x = 10,000, and c = 50,000, the equation becomes:

100,000 = 10,000^2 * a + 10,000 * b + 50,000

when y = 250,000 and x = 20,000, and c = 50,000, the equation becomes:

250,000 = 20,000^2 * a + 20,000 * b + 50,000

subtract 50,000 from both sides of each equation and you get:

50,000 = 10,000^2 * a + 10,000 * b
200,000 = 20,000^2 * a + 20,000 * b

multiply both sides of the first equation by 2 and you get:

100,000 = 2 * 10,000^2 * a + 2 * 10,000 * b
200,000 = 20,000^2 + a + 20,000 * b

simplify the first equation and leave the second equation as is to get:

100,000 = 2 * 10,000^2 * a + 20,000 * b
200,000 = 20,000^2 * a + 20,000 * b

subtract the first equation from the second equation to get:

100,000 = 20,000^2 * a - 2 * 10,000^2 * a

simplify to get:

100,000 = 200,000,000 * a

divide both sides of the equation by 200,000,000 to get:

100,000 / 200,000,000 = a

solve for a to get:

a = .0005

go back to the first original equation of 50,000 = 10,000^2 * a + 10,000 * b and replace a with .0005 and solve for b to get b = 0.

you have a = .0005 and b = 0 and c = 50,000

the general equation of y = ax^2 + bx + c becomes:

y = .0005x^2 + 0x + 50,000 which becomes:

y = .0005x^2 + 50,000










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