A variable line `x/a + y/b = 1` moves in such a way that the harmonic mean of a and b is 8. Then the least area of triangle made by the line with the

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A variable line `x/a + y/b = 1` moves in such a way that the harmonic mean of a and b is 8. Then the least area of triangle made by the line with the

A variable line `x/a + y/b = 1` moves in such a way that the harmonic mean of a and b is 8. Then the least area of triangle made by the line with the coordinate axes is (1) 8 sq. unit (2) 16 sq. unit (3) 32 sq. unit (4) 64 sq. unit
A. 8 sq. unit
B. 16 sq. unit
C. 32 sq. unit
D. 64 sq. unit

Monjur Alahi

9 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
Given line is `(x)/(a)+(y)/(b)=1`
It meets the axes at point A(a,0) and B(0,b).
Area of triangle AOB, `Delta = (1)/(2)` ab
Given that 8 is H.M. of a and b.
`therefore (1)/(a)+(1)/(b)=(1)/(4)`
`rArr b = (4a)/(a-4)`
`therefore Delta = (2a^(2))/(a-4)`
Differentiating w.r.t. a, we get
` (dDelta)/(da) = 2(2a(a-4)-a^(2))/((a-4)^(2))`
`rArr (dDelta)/(da) = (2a(a-8))/((a-4)^(2))`
`"If " (dDelta)/(da) = 0, "then " a =8`
This is the value of a for which area is minimum
`therefore Delta_(min.) = 32 " sq.units."`