Question

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) draw several diagrams to illustrate the situation, some short boxes with large bases and some tall boxes with small bases. find the volumes of several such boxes. (b) draw a diagram illustrating the general situation. let x denote the length of the side of the square being cut out. let y denote the length of the base. (c) write an expression for the volume v in terms of both x and y.

Answer 1

The largest volume that the box can have is two cubic feet.

• (a) Please see the attached graph created with MS Excel that show the variation of volume with the side length of the square cut out.

• (b) Please find attached the diagram created with MS Visio, illustrating the general situation.  

• (c) The expression for the volume v in terms of both x and y is V = y²·x.

Reasons:

Given parameters are;

The side length of the box = 3 ft.

x represents the side of the cut square, we have;

Height of the box = x

Width of the box, y = 3 - 2·x

length of the box, y = 3 - 2·x

The volume of the box, V = y²·x

Therefore; V = (3 - 2·x)²·x = 4·x³ - 12·x² + 9·x

At the largest volume, we have;

$\dfrac{dV}{dx} = \dfrac{d}{dx} \left(4 \cdot x^3 - 12 \cdot x^2 + 9 \cdot x\right) = 12 \cdot x^2 - 24 \cdot x + 9 = 0$

Which gives;

4·x² - 8·x + 3 = 0

$x = \dfrac{3}{2}$, or $x = \dfrac{1}{2}$

The value, $x = \dfrac{3}{2}$, gives the minimum volume of the box, given that $\dfrac{3}{2}$ is half the length of the cardboard, therefore;

y = 3 - 2·x

We get;

$y = 3 - 2 \times \dfrac{3}{2} = 3 - 3 = 0$

Which gives;

$V = 4 \cdot \left(\dfrac{1}{2} \right)^3 - 12 \cdot \left(\dfrac{1}{2} \right)^2 + 9 \cdot \left(\dfrac{1}{2} \right) = 2$

The largest volume that the box can have, $V_{max}$ is 2 ft.³

(a) The volume of boxes with their height, (side length of square being cut out) x  are plotted on the attached graph.

(b) Please find attached the drawings representing the general situation.

(c) From the question analysis, we have;

The volume of the box in terms of x and y is V = y²·x

The volume of the box in terms of x only is V = 4·x³ - 12·x² + 9·x

The volume of the box in terms of y only is  $V = y^2 \times \left(\dfrac{3 - y}{2} \right)$

Learn more here:

https://brainly.com/question/17983043