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The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

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प्रश्न

The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

योग
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उत्तर

Let us take either of the equal side AB and AC be x

⇒ `dx/dt = -3` cm/sec.

If A is the corresponding area of ΔABC

`A = 1/2` base × height = `1/2  b sqrt (AB^2 - BD^2)`

`= b/2 sqrt (x^2 - (b/2)^2)`

`= b/4 sqrt (4x^2 - b^2)`

`(dA)/dt = b/4 xx 1/2 xx (8x)/ sqrt (4x^2 - b^2)  dx/dt`

`= (bx)/ sqrt (4x^2 - b^2)  dx/dt`

`((dA)/dt)_(x = b) = ((b xxb)/ sqrt (4b^2 - b^2)) (-3)`

`= - sqrt(3b)` 

Area is decreasing at the rate of `b sqrt3` cm2 sec

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अध्याय 6: Application of Derivatives - Exercise 6.6 [पृष्ठ २४२]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.6 | Q 3 | पृष्ठ २४२

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