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March 1, 2015

March 1, 2015

Posted by **sadia khan** on Saturday, January 19, 2013 at 2:42am.

- maths -
**Reiny**, Saturday, January 19, 2013 at 8:53amlet each side of the cut-out square be x cm

where 0 < x < 12

base of box is (24-2x) by (45-2x)

height will be x

volume = x(24-2x)(45-2x)

= x(1080 - 138x + 4x^2) = 4x^3 - 138x^2 + 1080x

d(volume)/dx = 12x^2 - 276x + 1080

= 0 for a max/min of volume

x^2 - 23x + 90 = 0

(x-18)(x-5) = 0

x = 18 or x = 5

but x < 12 , so x = 5

a square of 5 by 5 cm should be cut out.

test:

if x = 4.8

volume = 4.8(24-9.6)(45-9.6) = 2446.848

if x = 5

volume = 5(24-10)(45-10) = 24050

if x = 5.1

volume = 5.1(24-10.2)(45-10.2) = 2449.224

max when x=5

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