# Calculus

posted by
**Tezuka** on
.

Find two numbers such that the sum of the first plus three times the second is 600 and their product is a maximum.

x + 3y=600

xy is a max. This is the problem statement.

let U be the function equal to xy.

U=xy = y(600-3y)

Now, maximize U (take derivative, set to zero), and solve for y.

I see BobP. set this up, but let me clarify a little.

Let x and y be the two number, as Bob had, then x+3y=600 and P=x*y. Since this is only one variable calculus, we need to express y in terms of x, so from

x+3y=600 we have y=(600-x)/3 = 200-(1/3)x

Now use that function of x for y in the product to get

P=x*[200-(1/3)x]=200x - (1/3)x^2

Find dP/dx, set to 0 and solve. You should also be in the habit of testing the endpoints and any other critical values too, sometimes these questions can throw a curve at you.