# Maximizing the volume of a box.

Suppose that the paper is 45 cm wide and 60 cm long. how much should you cut from the corners to form the box with maximum capacity, and what would be the width, length and volume of the resulting box?

Imagine cutting an $x$ by $x$ square out of each corner, and turning up the $4$ flaps to form a lidless box.

The height of the box will be $x$. The width will be $45-2x$, and the length $60-2x$. so the volume $V(x)$ of the box will be given by $$V(x)=x(45-2x)(60-2x).$$

We want to maximize $V(x)$, where $0\le x\le 45/2$.

To maximize, go through the usual calculus routine. First multiply out the terms in $V(x)$.

Added: Multiply out our expression for $V(x)$. We get $V(x)=4x^3-210x^2+2700x$. Thus $V'(x)=12x^2-420x+2700$. Set this equal to $0$ and solve for $x$. It is convenient but not necessary to first divide through by $12$. We get $x^2-35x+225=0$. Solve using the Quadratic Formula. We get $x=\frac{35\pm\sqrt{325}}{2}$. One of the roots is outside our interval. The endpoints $x=0$ and $x=45/2$ give $0$ volume. So volume is maximized when $x=\frac{35-\sqrt{325}}{2}$.

Now if we want to find the maximum volume, substitute the above value of $x$ in the expression for $V(x)$. The same applies to the length and width of the base. It seems sensible to use a decimal approximation to $x$ in these final calculations.

• The answer would be ? – Michael Rametta Mar 27 '13 at 3:23
• im still confused – Michael Rametta Mar 27 '13 at 3:24
• The volume is $V(x)=4x^3-210x^2+2700x$. Derivative is $12x^2-420x +2700$. Find out where this is $0$. Maybe use the Quadratic Formula. Maybe simplify first by dividing by $12$. Two roots, of which one is obviously no good. In principle need to check endpoints $0$ and $45/2$, but those are no good, they give volume $0$. – André Nicolas Mar 27 '13 at 3:35
• x=(-5/2)(7+sqrt(13)) and (5/2)(sqrt(13)-7) – Michael Rametta Mar 27 '13 at 3:39
• so what would be the size of the square we cut out, the length width, and volume? – Michael Rametta Mar 27 '13 at 3:41

Cut a length of 'x' from each end of the sheet and fold the sheet to make an open box.So the height would be $x$( the portion cut)

The measurements of this box would be :

length $= 60-2x$

breadth $=45-2x$

height $=x$

So volume of this box formed,V = $(60-2x)(45-2x)(x)$

Since we need to maximize this, differentiate V w.r.t x and find out the maxima. Substitute this maxima in the 'V' expression to get the maximum value