maths
posted by sadia khan .
an open box of rectangular base is to be made from 24 cm by 45cm cardboard by cutting out squares sheets of equal size from each corner and bending the sides.find the dimensions of the corner squares to obtain a box having largest volume.

let each side of the cutout square be x cm
where 0 < x < 12
base of box is (242x) by (452x)
height will be x
volume = x(242x)(452x)
= 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
(x18)(x5) = 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(249.6)(459.6) = 2446.848
if x = 5
volume = 5(2410)(4510) = 24050
if x = 5.1
volume = 5.1(2410.2)(4510.2) = 2449.224
max when x=5
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