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Question
A box with a square base is to have an open top. The surface area of the box is 192 sq cm. What should be its dimensions in order that the volume is largest?
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Solution
Let x cm be the side of square base and h cm be its height.
Then x2 + 4xh = 192
∴ h = `(192 - x^2)/(4x)` ...(1)
Let V = `x^2"h"`
= `x^2((192 - x^2)/(4x))` ...[By (1)]
∴ V = `(1)/(4)(192x - x^3)`
∴ `"dV"/("d"x) = (1)/(4) "d"/("d"x)(192x - x^3)`
= `(1)/(4)(192 xx 1 - 3x^2)`
= `(3)/(4)(64 - x^2)`
and
`("d"^2"V")/("d"x^2) - (3"d")/(4"d"x)(64 - x^2)`
= `(3)/(4)(0 - 2x)`
= `-(3)/(2)x`
For maximum V, `"dV"/("d"x)` = 0
∴ `(3)/(4)(64 - x^2)` = 0
∴ x2 = 64
∴ x = 8 ...[∵ x > 0]
and
`(("d"^2V)/("d"x^2))_("at" x = 8)`
= `-(3)/(2) xx 8`
= – 12 < 0
∴ by the second derivative test, V is maximum at x = 8.
If x = 8,
h = `(192 - 64)/(4(8)`
= `(128)/(32)`
= 4
Hence, the volume of the box is largest, when the side of square base is 8 cm and its height is 4 cm.
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