Question

Radioactive waste is to be disposed of in fully enclosed lead boxes of inner volume 200 cm^3. The base of the box has dimensions in the ratio 2:1.Explain why the inner surface area of the box is given by A(x)=4x^2 + 600/xcm^2

Answer 1

See v=lxbxh
l:b=2:1
thus l=2x b=x
v=2x x x x h
\[200=2x^{2}h\]
\[h=\frac{ 200 }{ 2x^{2}}\]
h=100/x2

Answer 2

now area of inner surface is
A(x)=2xlxb+2xbxh+2xlxh.
substitute l=2x b=x and h=100/x^2

Answer 3

A(x)=\[2\times 2x \times x +2\times x \times \frac{ 100 }{ x^{2} } + 2 \times 2x \times \frac{100}{x^{2}}\]

Answer 4

now just simplify..hope this helps

Answer 5

Thank you! This helps a lot.