A rectangular box with volume 320 ft^3 is built with a square base and top. The cost is 1.50 DOL

A rectangular box with volume 320 ft^3 is built with a...

A rectangular box with volume $320 \mathrm{ft}^{3}$ is built with a square base and top. The cost is 1.50 DOLLARS $/ \mathrm{ft}^{2}$ for the bottom, 2.50 DOLLARS $/ \mathrm{ft}^{2}$ for the sides, and 1 DOLLARS $/ \mathrm{ft}^{2}$ for the top. Let $x=$ the length of the base, in feet. (GRAPH CANNOT COPY). a) Express the cost of the box as a function of $x .$ b) Find the domain of the function. c) Graph the function with a graphing calculator. d) What dimensions minimize the cost of the box?

A rectangular box with volume $320 \mathrm{ft}^{3}$ is built with a square base and top. The cost is 1.50 DOLLARS $/ \mathrm{ft}^{2}$ for the bottom, 2.50 DOLLARS $/ \mathrm{ft}^{2}$ for the sides, and 1 DOLLARS $/ \mathrm{ft}^{2}$ for the top. Let $x=$ the length of the base, in feet.
(GRAPH CANNOT COPY).
a) Express the cost of the box as a function of $x .$
b) Find the domain of the function.
c) Graph the function with a graphing calculator.
d) What dimensions minimize the cost of the box?

Key Concepts

Cost Function Modeling

This concept involves creating a function that represents the total cost based on the dimensions of the object. It requires organizing the cost contributions from different components (such as the bottom, top, and sides) by multiplying their areas by the corresponding cost per unit area, then expressing these areas in terms of a single variable using relationships among the dimensions.

Volume Constraint

When the volume of an object is fixed, a volume constraint is used to relate the dimensions of the shape. In this context, the constraint enables one to express one dimension (such as the height) in terms of another (such as the base side length), thereby reducing the number of independent variables in the problem.

Domain Determination

This concept focuses on identifying the set of all possible values for the independent variable that are physically and mathematically viable. For geometric problems, the domain is restricted to values that yield positive dimensions and meaningful physical constructs, ensuring that the function is defined and realistic.

Graphical Analysis

Graphical analysis involves using tools like graphing calculators to visualize the behavior of a function. This helps in understanding the trends, locating the minimum or maximum points, and confirming that the calculated optimal values adhere to the problem’s constraints.

Optimization

Optimization is the process of finding the best possible solution—such as minimizing cost or maximizing efficiency—subject to given constraints. In this scenario, it involves using calculus techniques to determine the dimensions that minimize the overall cost of construction while satisfying the volume requirement.

Transcript

00:03 All right, so we have been given that we have a box that has a volume of 320 cubic feet.

00:19 And they give us certain information about this box.

00:22 We know that it costs $1 .50 per square foot for the material that's used to create the bottom of the box.

00:37 We know it costs $2 .50 per square foot for the material used for the sides, and it cost a dollar per square foot for the material used to make the top.

00:59 We know that this box we're trying to create has a square base, and we need to create a function that represents the cost to create this box.

01:20 So if the base, the sides of the base have dimensions of x, and it's a square base, so they both have to be x, right, x by x? and if i call the height y, then i can figure out a function here.

01:39 I'm going to start by talking about the area of each of these pieces, and then we'll talk about the total cost function.

01:47 So in other words, the area of each side, right? so we're talking about the sides of the box.

01:58 There's four sides.

01:59 And they all have the same dimensions.

02:00 They all have a dimension of x times y.

02:02 So if the area of one side is x times y, the area of all four sides is going to be four times that.

02:13 The bottom, length times width, x times x, it's going to have an area of x squared.

02:23 This was the area of the sides, of course, the bottom.

02:26 And the area of the top is also going to be x squared.

02:29 So if we talk about a cost function in terms of x, the sides, remember, were a dollar 50, nope, i'm sorry, the sides were 250 per square foot.

02:43 So it's going to be 250 times 4xy, plus to make the bottom is a $1 .50 times the, or $1 .50 per square foot.

02:57 So it's going to be $1 .50 times x squared.

02:59 And then the top is going to be 1 times x squared.

03:04 So this pretty much represents our cost function.

03:08 We have one issue, though.

03:10 250 times 4 is going to be 10 xy plus 150x squared plus 1x squared, which i could have combined like terms here.

03:23 So let's make this simpler.

03:25 We've got 10xy plus 250x squared.

03:31 The issue that we have here is we wanted a cost function in terms of x.

03:35 Notice i have this y here.

03:37 I've got two variables.

03:39 There's got to be a way to get rid of this variable.

03:43 And there is.

03:45 And that leads us back to the fact that they told us what the volume was at the beginning of the problem...

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