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26 Μαΐ 2020 · After cutting out the squares from the corners, the width of the open-top box will be 5-2x, and the length will be 7-2x. We’re being asked to maximize the volume of a box, so we’ll use the formula for the volume of a box, and substitute in the length, width, and height of the open-top box.
In this activity, students will work on a famous math problem exploring the volume of an open box. The aim is to create an open box (without a lid) with the maximum volume by cutting identical squares from each corner of a rectangular card.
If $1200\ \mathrm{cm}^2$ of material is available to make a box with a square base and an open top, find the largest possible volume of the box. The quantity we want to optimize is the volume of the box. Let $V$ be the volume of the box. We want to find the maximum value of $V$.
Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.
A rectangular box with a square base, an open top, and a volume of 216 216 in. 3 is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?
If 1200 $cm^2$ of material is available to make a box with a square base and an open top, find the largest possible volume of the box. So I setup my problem like this: $$ x^2 + 4xh = 1200 = P$$
Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.
