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Applications of Differentiation

Extracts from this document...

Introduction

Using differentiation to solve practical problems

1988 GCSE Maths Coursework Task

A rectangular sheet of card is 20cm long and 12cm wide. Equal squares are cut from each corner. The flaps are then folded to make an open box in the form of a cuboid, as shown below.

image00.png

12cm

                                                                                    20cm

  1. The box is to be filled with centimetre cubes. What size square should be cut from each corner so that the box will hold as many cubes as possible?
  2. The box is to be filled with sand. What size of square should now be cut from each corner so that the box will hold as much sand as possible?

image01.png

a)

...read more.

Middle

7.8

2.1

258.8

2.2 x 2.2

15.6

7.6

2.2

260.8

2.3 x 2.3

15.4

7.4

2.3

262.1

2.4 x 2.4

15.2

7.2

2.4

262.7

2.5 x 2.5

15.0

7.0

2.5

262.5

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)

...read more.

Conclusion

160;                                                                                    image03.pngimage03.png

At the stationary point, the rate of change of volume, with respect to the size of the square, is 0.i.e.image04.pngimage04.png

image05.pngimage05.pngimage06.pngimage06.png

This can be simplified to:

image07.pngimage07.png

We now have a quadratic we can solve for x.

𝒂=𝟑, 𝒃=−𝟑𝟐, 𝒄=𝟔𝟎        

Usingimage09.pngimage09.pngimage10.pngimage10.pngimage11.pngimage11.png

Therefore image12.pngimage12.pngorimage13.pngimage13.png

However, if x = 8.24 then the width of the cuboid would become negative, which is impossible. There for the only valid solution is:

image14.png

...read more.

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