Marginal cost with two variables, confused.

[nakota2k]

Homework Statement

Your weekly cost to manufacture x bicycles and y tricycles is

C(x,y)=20,000+60x+20y+50√(xy)

a. What is marginal cost of manufacturing a bicycle.

b. What is the marginal cost of manufacturing a tricycle.

c. What is the marginal cost of manufacturing a bicycle at a level of 10 bicycles and 10 tricycles?

d. What is the marginal cost of manufacturing a tricycle at a level of 8 bicycles and 8 tricycles?

Homework Equations

The Attempt at a Solution

Normally I thought I would derive then plug in for x to get marginal cost.
I derived x and got (25y^1/2)/x^(1/2) +60 but I don't know what to do next or if that is even right.

I derived y and got (25x^1/2)/(y^1/2) +20 again I don't know what to do next

 

[Ray Vickson]

Homework Statement

Your weekly cost to manufacture x bicycles and y tricycles is

C(x,y)=20,000+60x+20y+50√(xy)

a. What is marginal cost of manufacturing a bicycle.

b. What is the marginal cost of manufacturing a tricycle.

c. What is the marginal cost of manufacturing a bicycle at a level of 10 bicycles and 10 tricycles?

d. What is the marginal cost of manufacturing a tricycle at a level of 8 bicycles and 8 tricycles?

Homework Equations

The Attempt at a Solution

Normally I thought I would derive then plug in for x to get marginal cost.
I derived x and got (25y^1/2)/x^(1/2) +60 but I don't know what to do next or if that is even right.

I derived y and got (25x^1/2)/(y^1/2) +20 again I don't know what to do next

Yes, that is (roughly) what is meant by marginal cost.

For (a) and (b) that is all you can do, since a base pair (x,y) is not specified. However, in parts (c) and (d) you are given x and y values to work with. So, what do you think you should do next?

In principle, the true marginal cost of 1 bicycle, starting from a base of $(x_0,y_0)$, would be
\text{actual marginal cost } = C(x_0+1,y_0) - C(x_0,y_0)
For a cost function that does not "curve" too much (or for values of $x_0$ and $y_0$ that are not "too small" we have a reasonable approximation if we take
C(x_0+1,y_0) - C(x_0,y_0) \approx \frac{\partial C(x_0,y_0)}{\partial x},
so that is why I said "roughly" before. Basically, the partial derivative is what economists often use. To be absolutely sure of your usage, check the definitions in your textbook or course notes.

 

[nakota2k]

Ok this is what I got so far.

a. 60+25√y/x

b. 20+25√x/y

c. $85

d $45

 

[haruspex]

Ok this is what I got so far.

a. 60+25√y/x

b. 20+25√x/y

c. $85

d $45

Yes, but be careful with parentheses. Strictly speaking, √x/y means (√x)/y, but to get those numbers you must mean (correctly) √(x/y).