The figure shows an open box with a square base. The box is...

Name: Problem 31, Chapter 1: Precalculus
Uploaded: 2020-11-16T04:51:50-08:00
Duration: 2 min 47 s
Channel: Jorge Villanueva
Description: The figure shows an open box with a square base. The box is to have a volume of 10 cubic feet. Express the amount of material needed to construct the box, A , as a function of the length of a side of its square base, x

The figure shows an open box with a square base. The box is to have a volume of 10 cubic feet. Express the amount of material needed to construct the box, A , as a function of the length of a side of its square base, x

Key Concepts

Function Modeling

Function modeling involves expressing a physical quantity as a mathematical function of one or more variables. In this context, after performing the substitutions, the surface area is represented as a function of the side length of the square base. This function can be used for analyzing material requirements and optimizing the design of the box.

Variable Substitution

This concept refers to reducing the number of independent variables in a problem by using an established relationship, such as the volume constraint. By expressing the height in terms of the side length of the base, one substitutes back into the surface area formula. This substitution allows the material amount to be written as a function of a single variable, which is useful for further analysis like optimization.

Surface Area Expression

This key concept involves calculating the total area of all the material surfaces used in constructing a shape, excluding any surfaces that are open. For an open box, the surface area is calculated by combining the area of the bottom and the areas of the sides. The formula for surface area is central to minimizing or expressing the amount of material needed for construction.

Volume Constraint

This concept involves using the formula for the volume of a three?dimensional shape to relate its dimensions. In contexts like the open box with a square base, the volume is given by the product of the area of the base and the height. When the volume is fixed, this relationship allows one to express one variable (like the height) in terms of another (like the side length of the base).

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