suppose a company wants to introduce a new machine that will produce a rate of annual savings in dollars given by the function S'(x), where ex is the number of years of operation of the machine, while producing a rate of annual costs in dollars given by the function C'(x)
S’(x)=228-x^2, C’(x)= x^2+14/5x
To determine the optimal number of years of operation for the machine in order to maximize savings, we need to find the equilibrium point where savings equal costs.
The total savings over x years of operation is given by the antiderivative of the rate of savings function, S'(x):
S(x) = ∫(228 - x^2) dx
S(x) = 228x - x^3/3 + C1
And the total costs over x years of operation is given by the antiderivative of the rate of costs function, C'(x):
C(x) = ∫(x^2 + 14/5x) dx
C(x) = x^3/3 + 7/10x^2 + C2
At equilibrium, the total savings should equal the total costs:
S(x) = C(x)
228x - x^3/3 + C1 = x^3/3 + 7/10x^2 + C2
Now, we can solve for x by setting the two equations equal to each other:
228x - x^3/3 = x^3/3 + 7/10x^2
228x - x^3/3 = 3/10x^2
Rearranging and simplifying:
x^3 + 30x^2 - 684x = 0
Factor out the common factor x:
x(x^2 + 30x - 684) = 0
Since x cannot be negative, we can disregard the negative solution for x. We can find the value of x by solving the quadratic equation:
x^2 + 30x - 684 = 0
From here, we solve for x to find the optimal number of years of operation for the machine.