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. Let x denote the length of the side of the square being cut out. Let y denote the length of the base.(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. (Do this on paper. Your teacher may ask you to turn in this work.)(b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols. (Do this on paper. Your teacher may ask you to turn in this work.)(c) Write an expression for the volume V in terms of x and y.V =(d) Use the given information to write an equation that relates the variables. (Do this on paper. Your teacher may ask you to turn in this work.)(e) Use part (d) to write the volume as a function of x.V(x) =(f) Finish solving the problem by finding the largest volume that such a box can have.V = ft3

Answer:              Thus, maximum value of box is 2 square feet at [tex]x = \frac{1}{2}[/tex].Step-by-step explanation:A square piece of cardboard of side 3 feet is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (3-2x) and the width becomes (3-2x).Volume is given by V = [tex]V = Length\times Width\times Height\\V = (3 - 2x)^2x = x(4x^2-12x+9) = 4x^3-12x^2+9x\\So,\\V(x) = 4x^3-12x^2+9x[/tex]First, we differentiate V(x) with respect to x, to get,[tex]\frac{d(V(x))}{dx} = \frac{d(4x^3-12x^2+9x)}{dx} = 12x^2 - 24x +9[/tex]Equating the first derivative to zero, we get,[tex]\frac{d(V(x))}{dx} = 0\\\\12x^2 - 24x +9 = 0[/tex]Solving, with the help of quadratic formula, we get,[tex]x =\frac{3}{2}, \frac{1}{2}[/tex]Again differentiation V(x), with resopect to x, we get,[tex]\frac{d^2(V(x))}{dx^2} = 24x - 24[/tex]At x = [tex]\frac{1}{2}[/tex],[tex]\frac{d^2(V(x))}{dx^2} < 0[/tex]Thus, maxima occurs at x = [tex]\frac{1}{2}[/tex] for V(x).Thus, largest volume the box can have occurs when [tex]x = \frac{1}{2}[/tex].Maximum value of volume V([tex]\frac{1}{2}[/tex]) = [tex](3-1)^2\times \frac{1}{2} = 2\text{ square feet}[/tex]Thus, maximum value of box is 2 square feet at [tex]x = \frac{1}{2}[/tex].