Given R=\({12000\over12}=1000 units /month\)
C₁= 20 paise per unit / month
C₂= Rs.350 per run
(i) \(q_o=\sqrt{2C_3R\over C_1}=\sqrt{2\times 350\times 1000\over 0.20}=1870 units/ run\)
(ii)\(t_o=\sqrt{2C_3\over C_1R}=\sqrt{2\times 350\over 0.20\times 1000}=1.87 months\)
=1.87 x 30 = 56 days
(iii)\(C_o=\sqrt{2C_1 C_3R}=\sqrt{2\times 0.20\times12\times350\times(1000\times12)}\)
=Rs. 4490 per year

Given C =\({1\over5}x^2-6x+100\)
Differentiating w.r.t. 'x' we get,
\({dC\over dx}={1\over5}(2x)-6\)
\({dC\over dx}=0\)
\(\Rightarrow {2x\over 5}-6=0\)
\(\Rightarrow {2x\over 5}=6\)
\(\Rightarrow x={6\times5\over2}=15\)
Now  \({d^2C\over dx^2}={2\over5}>0\)
\(\therefore\)Cost function is minimum when x = 15.

Let y=x³-6x²+7
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=3x^2-12x\)
\({dy\over dx}=0\)
\(\Rightarrow3x^2-12x=0\)
\(\Rightarrow3x(x-4)=0\)
\(\Rightarrow x=0 \ or \ x=4\)
\({d^2y\over dx^2}=6x-12\)
when x=0 \({d^2y\over dx^2}=-12<0\)
\(\therefore \) y is maximum at x=0
\(\therefore \) maximum value=0³-6(0)²+7=7
when x=4, \({d^2y\over dx^2}=-6(4)-12=12>0\)
\(\therefore \) y is minimum at x=4
\(\therefore \) Minimum value =4⁴-6(4)²+7=64-96+7=-25
Hence maximum value is 7 and minimum value is -25.

Let y= 75-12x+6x²-x³
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=0-12+12x-3x^2=-3(x^2-4x+4)=-3(x-2)^2\)
\(\Rightarrow{dy\over dx}\le 0\) for all \(x \in (-\infty , \infty)\)
\(\therefore \) y decreases as x increases.

Given y=1+1/x
Differentiating w.r.t. 'x' we get,
\({dy\over dx}={-1\over x^2}<0\) for all real values of x, except x=0.
\(\therefore \) y is a strictly decreasing function for all real values of x(x\(\neq\)0).