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Dono murbruno 1911sihonemniz2A 1910D SUDOM Module 8 Cover Assignment Container Conundrum Total: 5 marks You have been given 1
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Answer #1

Case 1:

Let us consider a cuboid as below

here width = length = x,height = h

given surface area = 100 in^2

Total surface area = 2x^2 + 4xh

Max volume = x^2h

So,

2.1? + 4.ch = 100 4.ch = 100 -2.22 100 - 2.c 4.

Put h in Volume equation

volume = x^2h

2100 - 22 100 V = 25 -

b = 2.5 - 302,

100 – 20070)2 576 4(577) = 3 = 4.0825 3

So x and h are same so its a cube

Case 2:

Consider a cylinder as below:

2 ur A = h(2ar) - - - - - - L - A= 20$? + h(29tr)

h= 22 + 2xrh = 100 100 — 21:2 2

Volume=\pi r^2h

3 V = ar? __2,100 - 22 2 ) = 50r -

dv = 0 = 50 – 37r2 = 0 50 = 2.3033 T=1

h = 100 — 22 — = 4.6063 2ar =

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