Question 858241
we are given b= (t+3)^2 and h= t^2+3
Area of triangle is 1/2 * b * h
therefore
A = (1/2) * (t+3)^2 * (t^2+3)
A = (1/2) * (t^2+6t+9) * (t^2+3)
A = (1/2) * (t^4 + 6t^3 + 12t^2 + 18t + 27)
A = t^4/2 + 3t^3 + 6t^2 + 9t + 27/2
now
dA/dt = 2t^3 + 9t^2 + 12t + 9
for t = 3
dA/dt = 2*27 + 9*9 + 36 + 9 = 180