The two equal sides of an isosceles triangle with a fixed base 'b' cm are decreasing at the rate of 4 ~cm / sec . The rate at which the area is decreasing when each of the equal sides becomes equal to the base is

Step-by-step solution

Let the equal sides of the isosceles triangle be x and the base be b . The height h of the triangle is given by h = x^2 - (b/2)^2 = x^2 - b^2/4 . The area A of the triangle is A = 1 2 b h = b 2 x^2 - b^2 4 . Differentiating with respect to time t , we get dA dt = b 2 1 2 x^2 - b^2/4 2x dx dt = bx 2 x^2 - b^2/4 dx dt . Given dx dt = -4 cm/sec and we need the rate of change when x = b . Substituting x = b into the expression for dA dt : dA dt = b(b) 2 b^2 - b^2/4 (-4) = b^2 2 3b^2/4 (-4) = b^2 2(b 3 /2) (-4) = b^2 b