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anonymous
 3 years ago
FIND: The minimal average cost (ATTACHED.)
anonymous
 3 years ago
FIND: The minimal average cost (ATTACHED.)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0minimum is at the vertex compute \(\frac{b}{2a}\) with \(b=700,a=1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, i see you did that. hmmmm

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0weird i guess to minimize the cost, produce nothing

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i understood part d.... it's 280

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you found the vertex correctly, but you can't produce 350 items

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where on earth did the 280 come from?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's the production level that will minimize the average cost

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, i guess i have no idea what an average cost is

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1YOu have cost, to find average cost divide that by x. THEN take the derivative, that is marginal average cost. Minimize THAT.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i did c(X)/x, differentiated, then set it to zero. and got x=280

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why is that an "average cost"?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The minimal average cost?

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1Because C(x) gives the total cost of producing x items. So C(x)/x gives the average cost per item, at the production level x.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okaay but let's find part e please.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0learn something new every day

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry  I didn't mean to minimize the derivative... lol... I meant to minimize the average cost. Which you can do by setting the derivative of it =0. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's my main concern...

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1So you have average cost: A(x)=78400/x+700+x Take that derivative, set it = 0, and that's where your average cost is minimized.

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1OK, you good for e now?

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry  don't understand what?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0d) The production level that will minimize the average cost = 280. which is correct. e) The minimal average cost= ???

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1Just evaluate the average cost function at x=280

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1The average cost function: A(x)=78400/x+700+x which is just C(x)/x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then set it to zero right? @DebbieG

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1wha? no.... you found the production level that minimizes average cost already, by setting the derivative of average cost = 0, right? Now you just need to know what that average cost is  what is the average cost at that production level of x=280 So PLUG x=280 INTO the average cost function.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how about if i want to find the production level that will maximize profit.

DebbieG
 3 years ago
Best ResponseYou've already chosen the best response.1Then you need the profit function. Then find where it is maximized, by taking its derivative and set it = 0.
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