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The cost of manufacturing a rectangular box is as follows: The base costs 4 dollar per square foot, each of the sides cost 2 dollar per square foot and the top costs 1 dollar per square foot. Determine the cost function for the manufacture of this box. (Let $x$ be the length, $y$ the width and $z$ the height of the box.)

$C(x, y, z)=5 x y+4 x z+4 y z$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 1

Functions of Several Variables

Partial Derivatives

Johns Hopkins University

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

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.

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02:48

The cost of producing a re…

03:22

Construction Cost A rectan…

01:39

Consider the rectangular b…

03:27

The material for a closed …

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A rectangular box is to ha…

for this problem we are told that the cost of producing a rectangular boxes as follows. The sides cost $2 per square foot. The top $1 per square foot in the base. $4 per square foot were then told that if the volume of the boxes to be 10 cubic feet were asked to determine the dimensions that minimize the cost to begin. We'll have that. The cost will equal. Now we have the sides will be um One it is $2 per square foot. For the sides. We have $2 times. Now we have each side will be You have two times X. Y. for the front and back. You'll have two times Y. Z. For the other two sides. So we have two times two X. Y. Plus too easy for the sides. Then we have plus one times the cost of the top. The top is going to be X. Times E. And also we have that the base is $4 per square foot. So that will add on another four exit or plus five exit. So now we have our costs, we can multiply that too in so it's four X. Y. Plus four Y. Zed Plus five X. Set. And we're told that the volume must be 10 cubic feet. The volume is going to be X. Times Y. Times Z. And it must equal 10. So what we have to do here is figure out a way of minimizing the cost subject to the constraint X. Y. Z equals 10. So what we can do here is use the method of lagrange multipliers. So the first step is that we want to set the gradient of c. Radiant of C equal to lambda times the gradient of V. So we'll have first, when we take the partial derivative with respect to X, we'll have four Y plus five Z equals lambda times wise. It Then we have with respect to why we have four x plus four said equals lambda, X. Said. We then have uh with respect to Z. We have four Y plus five X equals lambda X. Y. And we also have our constraint X. Y. Z. equals 10. So we have a system of four equations with four unknowns one second here. So the solution that we get from that system of equations will be lambda equals for x equals two. Y equals five over to n. Z equals two.

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