Looking at the 1 x 1 square; if a 1 x 1 square is removed for the corners, the length becomes 20 -1 -1 =18cm. Similarly, for the width we have 12 -1 -1 =10cm. The height is the size of the square. The no. of cubes we can fit is the volume of the resulting cuboid i.e. 18(10)(1) = 180cm3 We can see that the maximum volume occurs when a 2 x 2 square is cut from each corner and hence the answer is “a 2x2 square.”
METHOD 1: Without differentiation
b) Similar to a) we can form a table and attempt some trial and error. However, the maximum volume will not be an integer value, since the sand will fill any space it is placed in. From the table in a) we can see the optimal square, which would give the largest value, is around a 2 x 2 cut out. The maximum value could have been reached before the 2 x 2 cut out, or occur between 2 x 2 and 3 x 3, so we must account for this, as we do not know where the maximum occurs yet.
Hence, cutting a 2.4 x 2.4 square from each corner yields the maximum volume of 262.7cm3 (1 d.p.)
METHOD 2: With differentiation
Consider the graph of the volume the cuboid against the size of the squares cut from the corners.
Volume
(Max)
Size of the sides of the squares cut out
The maximum value for the volume occurs at the stationary point for the curve. We can find this point using differentiation. To do this we need to form the equation for the curve, using the given information.
b) Let cm = the side of the square we will cut out.
From this the cuboid will have dimensions:
Length =
Width =
Height =
Therefore;
)
At the stationary point, the rate of change of volume, with respect to the size of the square, is 0. i.e.
This can be simplified to:
We now have a quadratic we can solve for x.
𝒂=𝟑, 𝒃=−𝟑𝟐, 𝒄=𝟔𝟎
Using
Therefore or
However, if x = 8.24 then the width of the cuboid would become negative, which is impossible. There for the only valid solution is:
