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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?
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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