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A producer sells two commodities, $x$ units of one commodity and $y$ units of the other. Suppose the Revenue derived from the sale of these two commodities is given by the equation $R(x, y)=12 x^{2} y^{3},$ and the cost function is given by $C(x, y)=x^{3} y^{3}+x^{2} y^{4}+5000,$ determine the maximum profit.

$\$ 1912$ at (4,6)

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 3

Extrema

Partial Derivatives

Harvey Mudd College

Baylor University

University of Nottingham

Idaho State University

Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

04:14

02:29

A company produces x units…

01:57

Find the maximum profit an…

02:18

The total-cost and total-r…

03:10

Suppose that the total re…

01:09

Suppose that the total rev…

04:55

Profit The profit from the…

02:50

for this problem we're told that a producer sells to commodities, X units of one commodity and Y units of the other. We were then told to suppose that the revenue derived from the sale of these two commodities is given by the equation are of X. Y equals 12 X squared Y cube. And the cost function is given by sea of X. Y equals x cubed Y cubed plus X squared white power four plus 5000. Were then asked to determine the maximum profit. So from the revenue and cost functions we first need to construct a profit function. Well that's going to be the revenue minus the cost. So it's going to be a 12 X squared y cubed minus X cubed y cube uh minus X Squared Whites Power 4 -5000. Now having that we then want to solve for when the gradient of the function It's going to equal 0 to find our critical parts. So our first element of the gradient from taking the partial derivative with respect to X is going to be 24 X Y cubed minus three X squared y cube -2. Xy is the power of four. And then with respect to why giving us our second component gives us 36 X squared y squared -3 x cubed y square -4 x squared y que now having that we have one solution where X equals zero, one solution where y equals zero. And both of those are you can imagine not exactly the case that we want to be looking at. In fact those will be minima And one more solution corresponding to the .46. So that gives us um the values that will give us the maximum profit. Then we just need to plug that in. So we want to find p of 46, which is going to come out to one moment here, That's going to equal 1,912.

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