Q.Case Study-I. A farmer has a piece of land. He observed that he got 600 units of fruits per tree by planting upto 25 trees and when 26 trees were grown, he received 15210 units of fruits, for 27 trees he ended up with 15390 fruits, for 28 trees he got 15540 fruits and this sequence of production of fruits continues in the same pattern as more trees, in excess of 25, were grown. Based on the above information answer the following questions:

1. If '$x$' more trees, in excess of 25 are grown, then the number of fruits produced per tree is

Modelling the yield as $600-15x$ fruits per tree for $x$ extra trees beyond 25 gives a total production $P(x)=(25+x)(600-15x)$, which is maximised by differentiating and setting $P'(x)=0$.

The data fix the model. Up to 25 trees each tree gives 600 fruits, and every extra tree beyond 25 reduces the yield of every tree by 15. So if $x$ more trees are planted beyond 25:

1. Fruits per tree $=600-15x$. (Check: $x=1\Rightarrow585$ and $26\times585=15210$; $x=2\Rightarrow570$ and $27\times570=15390$; $x=3\Rightarrow555$ and $28\times555=15540$ — matching the data.) → option (i).
2. Total production of the garden $=(\text{trees})\times(\text{fruits per tree})=(25+x)(600-15x)$ → option (iii).
3. Expanding, $P(x)=15000+225x-15x^2$, so the marginal production is $P'(x)=225-30x$ → option (ii).
4. The critical point solves $P'(x)=0\Rightarrow 225-30x=0\Rightarrow x=7.5$ → option (iii).
5. Since $x$ must be a whole number and $P''(x)=-30<0$ (a maximum), the two whole numbers either side of $7.5$ are tested: $P(7)=15000+1575-735=15840$ and $P(8)=15000+1800-960=15840$. Both give the same maximum, so $x=7$ or $8$, i.e. $25+x=$ 32 or 33 trees → option (ii).