# How to find the minimum length of wire required to construct the cuboid in this case?A piece of wire is used to construct a cuboid such that the length of its base is three times its width. The...

How to find the minimum length of wire required to construct the cuboid in this case?

A piece of wire is used to construct a cuboid such that the length of its base is three times its width. The condition is that the total surface area of the cuboid is 1274cm^2. How to find the minimum length of wire required to construct the cuboid?

jeew-m | College Teacher | (Level 1) Educator Emeritus

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If the base width is x cm then length will be 3x cm. lets assume the height is h cm.

If the length of the wire is L;

`L = (x+x)+(x+x)+(3x+3x)+(3x+3x)+(h+h+h+h)`

`L = 4(x+3x+h)`

`L = 4(4x+h)`

If the surface area is A;

`A = x*3x*2+x*h*2+3x*h*2 = 1274`

`6x^2+8xh = 1274`

` h = (1274-6x^2)/(8x)`

`L = 4(4x+h)`

`L = 16x+4*(1274-6x^2)/(8x)`

`L = 16x+(637-3x^2)/x`

For minimum or maximum L;

`(dL)/dx = 0`

(dL)/dx

`= 16+(x(-6x)-(637-3x^2))/x^2`

`= 16-(3x^2+637)/x^2`

`(dL)/dx = 0 `

`16-(3x^2+637)/x^2 = 0`

`16x^2 = 3x^2+637`

`13x^2 = 637`

`x = 49` since `x>0` .

When `x<49` say `x = 1` then `(dL)/dx <0`

When `x=49` then `(dL)/dx = 0 `

When `x>49` say `x = 50` then `(dL)/dx >0`

So `(dL)/dx ` is changing from `(-) rarr (0) rarr (+)`

So L has a minimum.

x = 49cm

h = 2.5 cm

L = `4(4*49+2.5)` = 794 cm

Minimum length of the wire 794 cm.

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