# A person wants to

The cost of planting trees is given by

C(x) = x3 - 45x2 + 600x where x is number of trees

We have to find the number of trees x such that the cost C(x) is minimum which means we have to find at what value of x we get minimum value for C(x)

Let us find where C(x) attains its minimum value

For this let us first identify where C(x) is strictly increasing/decreasing

We know that if C’(x) > 0 then C(x) is strictly increasing and if C’(x) < 0 then C(x) is strictly decreasing

Let us find C’(x)

C’(x) = 3x2 – 90x + 600

C’(x) = 3x2 – 30x – 60x + 600

C’(x) = 3x (x – 10) – 60(x – 10)

C’(x) = (3x – 60) (x – 10)

C’(x) = 3(x – 20) (x – 10)

C’(x) is 0 at x = 10 and x = 20

Now we will check whether C’(x) is greater than 0 or less than 0 in domain (-∞, 10), (10, 20) and (20, ∞) In (-∞, 10) C’(x) > 0 hence C(x) is strictly increasing in (-∞, 10)

In (10, 20) C’(x) < 0 hence C(x) is strictly decreasing in (10, 20)

In (20, ∞) C’(x) > 0 hence C(x) is strictly increasing in (20, ∞)

Observe that as C(x) is strictly decreasing in (10, 20) the graph of C(x) will be going downwards as one goes from 10 to 20 hence the minimum value of C(x) will be at 20

Hence the person should order for 20 trees to pay the minimum

As local authority has imposed a restriction that it can plant 10 to 20 trees in one community park 20 is a feasible number hence 20 people should order 20 trees

Now cost of 20 trees that is C(x) at x = 20 C (20)

C (20) = 203 – 45(20)2 + 600(20)

C (20) = 8000 – 45 × 400 + 12000

C (20) = 20000 – 18000

C (20) = Rs 2000

Hence the person should order 20 trees which will cost the minimum Rs 2000.

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