## hamidic 3 years ago a right angled triangle with length of the edges of the right angle "20" cm and"15"cm ;the first decrease by rate 2cm/sec . the second increase by 3cm/sec. find at any moment the rate of the change of the area and when the change stop

1. completeidiot

$\frac{dy}{dt} = 2 cm/\sec$ $\frac{dx}{dt} = 3 cm/\sec$ $determine \frac{dA}{dt}$

2. completeidiot

sorry wrote that wrong, should be -2cm/sec

3. hamidic

i need the area value at any instant in term of "t"

4. completeidiot

A(t) = (1/2)(20-2t)(15+3t)

5. completeidiot

$\frac{d}{dt}A(t)= \frac{d}{dt}\frac{1}{2}*(20-2t)(15+3t)$

6. hamidic

that right ; this problem was at my test and i want to check the second part "wiil the variation stop or not?'

7. completeidiot

it has to at t=10, because after this point, the area would be negative and negative areas dont exist (sort of)

8. completeidiot

i mean you might see negative areas when doing derivatives, but for this problem, i dont think negative areas count

9. hamidic

but area at any instant =0.5(20-2t)(15+3t) ;then area=$-3t ^{2}+15t+150$ then ;the rate of change of area at any instant=-6t+15; at rate stops -6t+15=0;then the variation stops at (15/6) not at 10 ;if i have any mistake tell me please

10. y2o2

@completeidiot He asked when will the variation stop, so the idea is about letting the derivative(dA/dt) = 0 , not the the area [A(t)]= 0