Given condition:
p – [(1/2) x^2] = 48 ------------------------(1)
By taking differentiation of equation (1); dp/dt represents the decreasing rate of change in the tire price and dx/dt represents rate of change of the tires being introduced.
x = 9
p = 88.5
Tire price decreasing rate = dp/dt = 4$ per week
Solution:
Differentiation is performed implicitly.
dp/dt – [(1/2) d(x^2)/dt ]= d(48) /dt
Putting value of dp/dt;
– 4 – [(1/2) * 2 x (dx/dt)] = 0
Where “– “is due to decreasing rate of change of tires price and derivate of a constant “48” is equal to “0”
x dx/dt = -4
dx/dt = -4/x
Where x = 9
So dx/dt = -4/9
Hence the tire supply rate is decreasing the at rate of -4/9.