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Question
The perimeter of a triangle is 10 cm. If one of the side is 4 cm. What are the other two sides of the triangle for its maximum area?
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Solution
Let ABC be the triangle such that the side BC = a = 4 cm. Also, the perimeter of the triangle is 10 cm.
i.e. a + b + c = 10
∴ 2s = 10
∴ s = 5
Also, 4 + b + c = 10
∴ b + c = 6
∴ b = 6 – c
Let Δ be the area of the trangle.
Then Δ = `sqrt(s(s - a)(s - b)(s - c)`
= `sqrt(5(5 - 4)(5 - 6 + c)(5 - c)`
= `sqrt(5(c - 1)(5 - c)`
∴ Δ2 = 5(c – 1)(5 – c)
= 5(5c – c2 – 5 + c)
∴ Δ2 = 5(– c2 + 6c – 5)
Differentiable both sides ww.r.t. c, we get
`2Δ(dΔ)/(dc) = 5d/"dc"(-c^2 ++ 6c - 5)`
= 5 (– 2c + 6 x 1 – 0)
= 5 (– 2c + 6)
∴ `(dΔ)/"dc" = (5(-c + 3))/Δ`
and
`(d^2Δ)/(dc^2) = 5d/"dc"((-c + 3)/Δ)`
= `5.(Δd/"dc"(– c + 3) – ( – c + 3)(dΔ)/"dc")/Δ^2`
= `5.(Δ(– 1 + 0) – ( – c + 3)(dΔ)/"dc")/Δ^2`
= `5/Δ^2(-Δ - (c + 3)(dΔ)/"dc")`
= `(-5)/Δ^2[Δ + (c + 3)(dΔ)/"dc"]`
For maximum Δ, `(dΔ)/"dc"` = 0
∴ `(5( - c + 3))/Δ` = 0
∴ – c + 3 = 0 ...[∵ Δ ≠ 0]
∴ c = 3
If c = 3,
Δ = `sqrt(5(3 - 1)(5 - 3)`
= `2sqrt(5)`
∴ `((d^2Δ)/(dc^2))_("at" c = 3)`
= `(-5)/(4 xx 5)[2sqrt(5) + (3 + 3)(0)]`
= `sqrt(5)/(2) < 0`
∴ by the second derivative test, Δ is maximum when c= 3.
When c = 3, b = 6 – c = 6 – 3 = 3
Hence, the area of the triangle is maximum when the other two sides are 3cm and 3cm.
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