Question

A person wants to plant some trees in his community park. The local nursery has to perform this task. It charges the cost of planting trees by the following formula
\[f(x)={{x}^{3}}-45{{x}^{2}}+600x\], where x is the number of trees and f(x) is the cost of planting x trees in rupees. For how many trees should the person place the order so that he has to spend the least amount? Use calculus to answer these questions. Which value is being exhibited by a person?

Answer 1

Hint: Here in this question we are given a problem statement and we are asked to find the minimum number of trees required therefore we can solve it by using the method we find the minimum of a function. The function is given in the question. We can start by differentiating the function and after it we need to equate it to 0, on doing that we find the points or values at 0 where the function can be minimum or maxima. Then after finding those values we need to find the second derivative and check values (the values are the points for which the first derivative is 0), if the value for x in second derivative is positive the value refers to minimum and if the value for x is negative then it refers to the maximum of the function.

Complete step-by-step answer:
Now we are given the function whose minimum we need to find;
\[f(x)={{x}^{3}}-45{{x}^{2}}+600x\]
Now we take the differentiation of this function to further solve this question
\[f'(x)=3{{x}^{2}}-90x+600\]
Here to now find the values of x we equate it to zero
\[3{{x}^{2}}-90x+600=0\]
Dividing the equation by \[3\]
\[{{x}^{2}}-30x+200=0\]
Now solving this for x
\[(x-20)(x-10)=0\]
Now we know that
\[x=10,20\]
Taking the second differentiation
\[f''(x)=2x-30\]
Putting both values of x to find if we will get minimum at the point or maximum
\[f''(x)=2(10)-30=-10\] ; Here we will get maximum
\[f''(x)=2(20)-30=10\] ; Here we will get minimum
Therefore substituting \[20\] in the function we get
\[8000-18000+12000=2000\]
Therefore this is the least amount

Note: If we find that the second derivative is \[0\] then we can say that the value where we found it to be \[0\] is the point of inflection and the second derivative changes its sign at that point. To verify our answer, we can substitute the other value \[10\] in the given equation.