Question

A box is made of a piece of metal sheet 24cms square by cutting small squares from each corner and turning up the edges. If the volume of the box is maximum. Then the dimensions of the box are.
$\begin{align}
  & a)16,16,4 \\
 & b)9,9,6 \\
 & c)8,8,8 \\
 & d)9,9,8 \\
\end{align}$

Answer 1

Hint: Now we are given with a square piece of length 24cm. let us say that we cut squares from each corner of length a cm. Now the new length and breadth of the box will be 24 – 2a and the height will be a cms. We know that the volume of a box is $l\times b\times h$ . Hence we will use this formula to first find the equation from volume. Now we know to find the maximum we derivate and equate the function to zero. Hence we get conditions for extrema. We will check the function at both obtained values and determine which condition suggests maximum value.

Complete step by step solution:
Now we are given with the side of the square as 24cm.
Now we let us say that we cut small squares of length a cm from each corner.
Let us draw a figure of the following.

Hence the remaining length of the square is 24 – 2a.
Similarly the width of the rectangle will be 24 – 2a.
Now after folding we can say that the height of box will be a cms
Now we know that the volume of the box is given by $l\times b\times h$ where l is the length of box, b is the width of box and h is the height of box.
Now the volume of box is
$\Rightarrow V={{\left( 24-a \right)}^{2}}a$
Now we want the volume to be maximum.
Hence let us differentiate the volume and equate it to zero to find the condition for maxima.
Hence the maxima is at.
Now we know that $\dfrac{d\left( u.v \right)}{dx}=u'v+v'u$
$\begin{align}
  & \Rightarrow \dfrac{dV}{da}={{\left( 24-a \right)}^{2}}+\left( -4 \right)\left( 24-2a \right)a \\
 & \Rightarrow {{\left( 24-2a \right)}^{2}}=4\left( 24-2a \right)a \\
 & \Rightarrow {{\left( 24-2a \right)}^{2}}-4\left( 24-2a \right)a=0 \\
 & \Rightarrow \left( 24-2a \right)\left[ 24-2a-4a \right]=0 \\
 & \Rightarrow \left( 24-2a \right)\left[ 24-6a \right]=0 \\
\end{align}$
Hence either 24 – 2a = 0 or 24 – 6a = 0.
Hence we get either a = 12 and a = 4.
Now at a = 12 we have 24 – 2a = 0.
Hence we get length and breadth as 0.
Hence this is the condition form minimum.
Now similarly if a = 4 we have 24 – 2a = 16.
Now length = breadth = 24 – 2a and height = a
Hence the length breadth and height are 16, 16 and 4.
Hence the dimensions for maximum volume are 16, 16, 4.

So, the correct answer is “Option a”.

Note: Now note that when we have the equation ${{\left( 24-2a \right)}^{2}}=4\left( 24-2a \right)a$ we can cancel $24-2a$ on both sides and get the answer a = 12. But since the equation is a quadratic equation it must have two roots. Now always remember when we divide the equation by 24 – 2a which is nothing but cancelation we assume that $24-2a\ne 0$ . Hence the other case can be 24 – 2a = 0.
This will give us a second root which is a = 4.