Saturday

February 28, 2015

February 28, 2015

Posted by **Tezuka** on Saturday, September 30, 2006 at 10:19am.

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.

**Answer this Question**

**Related Questions**

Calculus - Find two positive numbers that satisfy the requirements: "The product...

calculus - Find two positive numbers that satisfy the given requirements: The ...

Calculus - Find two positive numbers that satisfy the given requirements. The ...

Math - Compute the maximum product for two positive numbers x and y with the ...

Math - Can someone tell me how to solve my problem? Do I just start plugging in ...

calculus - find two positive integers such that the sum of the first number and ...

Algebra - three numbers are such that the second is the difference of three ...

Calculus Optimization Problem - Find two positive numbers whose sum is 15 such ...

math-word problems with quadratics - I have 5 questions... 1) The sum of two ...

Math(Urgent) - Compute the maximum product for two positive numbers x and y with...