Question

An open box is to be made out of a piece of a square card board of sides 18cm by cutting off equal squares from the corners and turning up the sides. Find the maximum volumes of the box

Answer 1

Hint: We have to cut equal squared from each corner and have to maximize the area; we need to differentiate the quantity that we need to maximize to obtain the independent variable and again double differentiate the corresponding variables.

Complete step-by-step answer:
We have to cut equal squared from each corner and have to maximize the area
So let the length of side of the square which is to be cut from each corner be x cm
Now an open cubical box is formed by folding up the flag whose sides will be 18 – 2x, 18 – 2x and x cm respectively.
Volume of this cubical box can be written as
V = (18 – 2x)(18 – 2x)x
Now we need to maximize the volume so differentiate it with respect to x and put it equal to 0 to find x
$\dfrac{{dV}}{{dx}} = 0$
$\dfrac{{dV}}{{dx}} = 324 - 114x + 12{x^2} = 0$
So taking 12 common from above then we are having a quadratic equation
${x^2} - 12x + 27 = 0$
Factoring it we will get
${x^2} - 3x - 9x + 27 = 0$
\[x\left( {x - 3} \right) - 9\left( {x - 3} \right) = 0\]
Hence we are getting two values of x that is
X=3, 9
Now if we talk about x=9, then the dimension of open cuboids that are 18 – 2x, 18 – 2x and x will become 0, 0, 9 which is not possible hence we have only one value of x that is
X = 3
Now in order to be sure that value corresponding to max volume we have to ensure $\dfrac{{{\partial ^2}V}}{{\partial {x^2}}} < 0$ for this x
$\dfrac{{{\partial ^2}V}}{{\partial {x^2}}} = - 144 + 72 = - 72$ Clearly this is <0
Hence for x=3 we get max volume and its value is =
\[V = \left( {18-2x} \right)\left( {18-2x} \right)x = 15 \times 15 \times 3 = 675c{m^2}\]

Note: Whenever we need to find this type of problem we need to differentiate the quantity that we need to maximize to obtain the independent variable then simply verify when that value corresponds to max or min or not by finding a double derivative.