We have, A₁: A₂ :A₃ = 4: 4: 2
\(\therefore P\left(A_{1}\right)=\frac{4}{10}, P\left(A_{2}\right)=\frac{4}{10} \text { and } P\left(A_{3}\right)=\frac{2}{10}\) 
where A_I',A₂ and A₃ denote the event of choosing flower seeds A_1, A₂ and A₃ respectively,
Let E be the event that a seed germinates and E be the event that a seed does not germinate.
Then,\(P\left(\frac{E}{A_{1}}\right)=\frac{45}{100}, P\left(\frac{E}{A_{2}}\right)=\frac{60}{100}, P\left(\frac{E}{A_{3}}\right)=\frac{35}{100}\) 
and \(P\left(\frac{\bar{E}}{A_{1}}\right)=\frac{55}{100}, P\left(\frac{\bar{E}}{A_{2}}\right)=\frac{40}{100}, P\left(\frac{\bar{E}}{A_{3}}\right)=\frac{65}{100}\) 
(i) Probability that a randomly chosen seed to germinate,
\(P(E)=P\left(A_{1}\right) \cdot\left(\frac{E}{A_{1}}\right)+P\left(A_{2}\right) \cdot P\left(\frac{E}{A_{2}}\right)+P\left(A_{3}\right) \cdot P\left(\frac{E}{A_{3}}\right)\) 
\(=\frac{4}{10} \times \frac{45}{100}+\frac{4}{10} \times \frac{60}{100}+\frac{2}{10} \times \frac{35}{100}\) 
\(=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=\frac{490}{1000}=0.49\) 
\(\text { (ii) } P\left(\frac{\bar{E}}{A_{3}}\right)=1-P\left(\frac{E}{A_{3}}\right)=1-\frac{35}{100}=\frac{65}{100}\) 
\(\text { (iii) } P\left(\frac{A_{2}}{\bar{E}}\right)\) 
\(=\frac{P\left(A_{2}\right) \cdot P\left(\frac{\bar{E}}{A_{2}}\right)}{P\left(A_{1}\right) \cdot P\left(\frac{\bar{E}}{A_{1}}\right)+P\left(A_{2}\right) \cdot P\left(\frac{\bar{E}}{A_{2}}\right)+P\left(A_{3}\right) \cdot P\left(\frac{\bar{E}}{A_{3}}\right)}\) 
\(=\frac{\frac{4}{10} \times \frac{40}{100}}{\frac{4}{10} \times \frac{55}{100}+\frac{4}{10} \times \frac{40}{100}+\frac{2}{10} \times \frac{65}{100}}\) 
\(=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}}=\frac{\frac{160}{1000}}{\frac{510}{1000}}=\frac{16}{51}\) 
= 0.313725 = 0.314

Equation of plane passing through the intersection of given planes is
\(\vec{r} \cdot[(\hat{i}+3 \hat{j})+\lambda(3 \hat{i}-\hat{j}-4 \hat{k})]=6+0 \cdot \lambda\) 
\(\Rightarrow \vec{r} \cdot[(1+3 \lambda) \hat{i}+(3-\lambda) \hat{j}+\hat{k}(-4 \lambda)]=6\)  ...(i)
On dividing both sides by
\(\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}\) ,We get 
\(\frac{\vec{r} \cdot[(1+3 \lambda) \hat{i}+(3-\lambda) \hat{j}+\hat{k}(-4 \lambda)]}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}\) 
\(=\frac{6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}\) 
Given the perpendicular distance from origin is unity.
\(\therefore \frac{6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=1\) 
\(\Rightarrow (1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}=36\) 
\(\Rightarrow 1+9 \lambda^{2}+6 \lambda+9+\lambda^{2}-6 \lambda+16 \lambda^{2}=36\) 
\(\Rightarrow 26 \lambda^{2}+10=36 \Rightarrow \lambda^{2}=1\) 
\(\Rightarrow \lambda=\pm 1\) 
On putting the value of \(\lambda\) in Eq. (i), the required equation of plane are
\(\vec{r} \cdot[(1 \pm 3) \hat{i}+(3 \mp 1) \hat{j}+(\mp 4) \hat{k}]=6\) 
\(\Rightarrow \vec{r} \cdot[(1+3) \hat{i}+(3-1) \hat{j}+(-4) \hat{k}]=6\) 
and \(\vec{r} \cdot[(1-3) \hat{i}+(3+1) \hat{j}+4 \hat{k}]=6\) 
\(\Rightarrow \vec{r} \cdot(4 \hat{i}+2 \hat{j}-4 \hat{k})=6\) 
and \(\vec{r} \cdot(-2 \hat{i}+4 \hat{j}+4 \hat{k})=6\) 
and \(-2 x+4 y+4 z-6=0\) 

\(\frac{15}{2} \text { sq units }\)

\(\text {Required area }=2\left[\int_{0}^{2 a} \sqrt{6 a x} d x+\int_{2 a}^{4 a} \sqrt{16 a^{2}-x^{2}} d x\right]\) 
 
\(\frac{4}{3}(\sqrt{3}+4 \pi) a^{2} \text { sq units }\)

Let \(I=\int_{0}^{\pi} x \log |\sin x| d x\)   ...(i)  
\(\Rightarrow I=\int_{0}^{\pi}(\pi-x) \log |\sin (\pi-x)| d x\) 
\(=\int_{0}^{\pi}(\pi-x) \log |\sin x| d x\)   ..(ii)
On adding Eqs. (i) and (ii), we get
\(2 I=\pi \int_{0}^{\pi} \log |\sin x| d x\) 
\(\Rightarrow 2 I=2 \pi \int_{0}^{\pi / 2} \log |\sin x| d x\)    ...(iii)
\(\left[\because \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x, \text { if } f(2 a-x)=f(x)\right]\) 
\(\Rightarrow I=\pi \int_{0}^{\pi / 2} \log |\sin x| d x\)  ...(iv)
\(\Rightarrow I=\pi \int_{0}^{\pi / 2} \log |\sin (\pi / 2-x)| d x\) 
\(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) 
\(=\pi \int_{0}^{\pi / 2} \log |\cos x| d x\)   ...(v)
On adding Eqs. (iv) and (v), we get
\(2 I=\pi \int_{0}^{\pi / 2}(\log |\sin x|+\log |\cos x|) d x\) 
\(\Rightarrow 2 I I=\pi \int_{0}^{\pi / 2} \log |\sin x \cos x| d x\) 
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2} \log \left|\frac{2 \sin x \cos x}{2}\right| d x\) 
[multiply by 2 from numerator and denominator]
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2}(\log |\sin 2 x|-\log 2) d x\) 
\(2 I=\pi \int_{0}^{\pi / 2} \log |\sin 2 x| d x-\pi \int_{0}^{\pi / 2} \log 2 d x\) 
\(\Rightarrow 2 I=\pi \int_{0}^{\pi / 2} \log |\sin 2 x| d x-\pi \log 2[x]_{0}^{\pi / 2}\) 
Now, put \(2 x=t \Rightarrow d x=\frac{1}{2} d t\) 
Lower limit When \(x=0, \text { then } t=0\) 
Upper limit When \(x=\frac{\pi}{2}, \text { then } t=\pi\) 
\(\therefore \ 2 I=\frac{\pi}{2} \int_{0}^{\pi} \log |\sin t| d t-\frac{\pi^{2}}{2} \log 2\) 
\(\Rightarrow 2 I=\frac{\pi}{2} \int_{0}^{\pi} \log |\sin x| d x-\frac{\pi^{2}}{2} \log 2\) 
\(\Rightarrow 2 I=I-\frac{\pi^{2}}{2} \log 2\) [from ii]
\(\therefore I=-\frac{\pi^{2}}{2} \log 2=\frac{\pi^{2}}{2} \log \left(\frac{1}{2}\right)\)

\(I=\int e^{-3 x} \cos ^{3} x d x\) 
\(I=\int e^{-3 x}\left[\frac{\cos 3 x+3 \cos x}{4}\right] d x\) 
\(=\frac{1}{4}\left[\int e^{-3 x} \cos 3 x d x+3 \int e^{-3 x} \cos x d x\right]\) 
\(\begin{aligned} I=\frac{e^{-3 x}}{24}[\sin 3 x-\cos 3 x] -\frac{9}{40} e^{-3 x} & \cos x+\frac{3}{40} e^{-3 x} \sin x+C \end{aligned}\)

Let  \(I=\int \frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x} d x=\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{1-2\left(2 \cos ^{2} \frac{3 x}{2}-1\right)} d x\) 
\(\left[\begin{array}{l} \because \cos C+\cos D=2 \cos \frac{C+D}{2} \cdot \cos \frac{C-D}{2} \\ \text { and } \cos x=2 \cos ^{2}\left(\frac{x}{2}\right)-1 \end{array}\right]\) 
 \( \Rightarrow I=\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{3-4 \cos ^{2} \frac{3 x}{2}} d x=-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2}}{4 \cos ^{2} \frac{3 x}{2}-3} d x \)
\(=-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2}}{4 \cos ^{3} \frac{3 x}{2}-3 \cos \frac{3 x}{2}} d x \) 
\(\left[\text { multiply and divide by } \cos \frac{3 x}{2}\right]\) 
\( =-\int \frac{2 \cos \frac{9 x}{2} \cdot \cos \frac{x}{2} \cdot \cos \frac{3 x}{2}}{\cos 3 \cdot\left(\frac{3 x}{2}\right)} d x \)
\(=\left[\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\right] \) 
\( =-\int 2 \cos \frac{3 x}{2} \cdot \cos \frac{x}{2} d x \)
\(=-\int\left\{\cos \left(\frac{3 x}{2}+\frac{x}{2}\right)+\cos \left(\frac{3 x}{2}-\frac{x}{2}\right)\right\} d x \)
\(=-\int_{0}[(\cos 2 x+\cos x) d x \)
\(=-\left[\frac{\sin 2 x}{2}+\sin x\right]+C=-\frac{1}{2} \sin 2 x-\sin x+C \)

Let two men start from the point C with velocity veach at the same time.
Also, \(\angle B C A=45^{\circ}\) 

Since, A and B are moving with same velocity v, so they will cover same distance in same time 
Therefore, \(\Delta A B C\) is an isosceles triangle with AC = BC
Now, draw \(C D \perp A B\) 
Let at any instant t, the distance between them is AB
Let \(A C=B C=x \text { and } A B=y\) 
In \(\Delta A C D \text { and } \Delta D C B\) we get
\( \angle C A D=\angle C B D \)
\( \angle C D A=\angle C D B=90^{\circ} \)
\(\therefore \ \angle A C D=\angle D C B \text { or } \angle A C D=\frac{1}{2} \times \angle A C B\) 
\(\Rightarrow \angle A C D=\frac{1}{2} \times 45^{\circ} \Rightarrow \angle A C D=22.5^{\circ} \)
\(\therefore \sin 22.5^{\circ}=\frac{A D}{A C} \)
\(\Rightarrow \sin 22.5^{\circ}=\frac{y / 2}{x} \)
\(\Rightarrow \frac{y}{2}=x \sin 22.5^{\circ} \)
\(\Rightarrow y=2 x \cdot \sin 22.5^{\circ} \)
On differentiating both sides w. r. t. t, we get 
\( \frac{d y}{d t} =2 \cdot \sin 22.5^{\circ} \frac{d x}{d t}=2 \cdot \sin 22.5^{\circ} v \left[\because v=\frac{d x}{d t}\right][1] \)
\(=2 v \cdot \frac{\sqrt{2-\sqrt{2}}}{2}=v \sqrt{2-\sqrt{2}}\left[\because \sin 22.5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2}\right] \) 
which is the required rate at which A and B are being separated.

We have, \(A=\left[\begin{array}{rrr}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]\)
\(\therefore \quad|A|=1(-3)-2(-2)+0=1 \neq 0\)
Now, \(A_{11}=-3, A_{12}=2, A_{13}=2\),
\(A_{21}=-2, A_{22}=1, A_{23}=1, A_{31}=-4, A_{32}=2\)
and \(A_{33}=3\)

Also, we have the system of linear equations as
x-2 y =10 
2 x-y-z  =8 and -2 y+z=7
\( \therefore \quad \operatorname{adj}(A)=\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]^T=\left[\begin{array}{rrr} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right]\)
\(\therefore \quad A^{-1}=\frac{\operatorname{adj} A}{|A|} \)
\(=\frac{1}{1}\left[\begin{array}{rrr} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right]\)
\(\Rightarrow \quad A^{-1}=\left[\begin{array}{rrr} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right]\)
Also, we have the system of linear equations as  x - 2y = 10,,2x - y- z = 8 and - 2y +z = 7
In the form of CX = D
\(\left[\begin{array}{rrr} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 10 \\ 8 \\ 7 \end{array}\right]\)
where, \(C=\left[\begin{array}{rrr}1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\) and \(D=\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]\)
We know that \(\left(A^T\right)^{-1}=\left(A^{-1}\right)^T\)
\( \therefore \quad C^T=\left|\begin{array}{rrr} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right|=A\)
\(\therefore \quad X=C^{-1} D\) [using Eq. (i)]
\(\Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right] =\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right] \)
\( =\left[\begin{array}{c} -30+16+14 \\ -20+8+7 \\ -40+16+21 \end{array}\right]=\left[\begin{array}{r} 0 \\ -5 \\ -3 \end{array}\right]\)
\(\therefore x=0, y=-5 \text { and } z=-3\)

To prove,\(\left|\begin{array}{ccc} (b+c)^{2} & a^{2} & a^{2} \\ b^{2} & (c+a)^{2} & b^{2} \\ c^{2} & c^{2} & (a+b)^{2} \end{array}\right|=2 a b c(a+b+c)^{3}\) 
\(\mathrm{LHS}=\left|\begin{array}{ccc} (b+c)^{2} & a^{2} & a^{2} \\ b^{2} & (c+a)^{2} & b^{2} \\ c^{2} & c^{2} & (a+b)^{2} \end{array}\right|\) 
\(\text { Applying } C_{2} \rightarrow C_{2}-C_{1} \text { and } C_{3} \rightarrow C_{3}-C_{1}, \text { we get }\) 
\(=\left|\begin{array}{ccc} (b+c)^{2} & a^{2}-(b+c)^{2} & a^{2}-(b+c)^{2} \\ b^{2} & (c+a)^{2}-b^{2} & 0 \\ c^{2} & 0 & (a+b)^{2}-c^{2} \end{array}\right|\) 
\(=\left|\begin{array}{ccc} (b+c)^{2} & (a+b+c)(a-b-c) & (a+b+c)(a-b-c) \\ b^{2} & (c+a+b)(c+a-b) & 0 \\ c^{2} & \mathbf{O} \quad 0 & (a+b+c)(a+b-c) \end{array}\right|\) 
Taking (a + b + c) common from C₂ and C₃, we get
\(=(a+b+c)^{2}\left|\begin{array}{ccc} b^{2}+c^{2}+2 b c & a-b-c & a-b-c \\ b^{2} & c+a-b & 0 \\ c^{2} & 0 & a+b-c \end{array}\right|\) 
\(\text { Applying } R_{1} \rightarrow R_{1}-\left(R_{2}+R_{3}\right) \text { , we get }\)
\(=(a+b+c)^{2}\left|\begin{array}{ccc} 2 b c & -2 c & -2 b \\ b^{2} & c+a-b & 0 \\ c^{2} & 0 & a+b-c \end{array}\right|\)
\(\text { Applying } C_{2} \rightarrow C_{2}+\frac{1}{b} C_{1} \text { and } C_{3} \rightarrow C_{3}+\frac{1}{c} C_{1}, \text { we get }\) 
\(\mathrm{LHS}=(a+b+c)^{2}\left|\begin{array}{ccc} 2 b c & 0 & 0 \\ b^{2} & c+a & \frac{b^{2}}{c} \\ c^{2} & \frac{c^{2}}{b} & a+b \end{array}\right|\) 
Expanding along R₁ we get
\( \mathrm{LHS} =(a+b+c)^{2}[2 b c\{(c+a)(a+b)-b c\}-0+0] \)
\(=2 b c(a+b+c)^{2}\left[a c+a^{2}+a b+b c-b c\right] \)
\( =2 a b c(a+b+c)^{3}=\mathrm{RHS} \)

Let \(\Delta=\left|\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \end{array}\right|\) 
[applying R_1➝ R₁ + R₂ + R₃]
\(=(a+b+c)\left|\begin{array}{lll} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{array}\right|\) 
\(=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ b-a & c-a & a \\ c-b & a-b & b \end{array}\right|\) 
\(\left[\text { using } C_{1} \rightarrow C_{1}-C_{3} \text { and } C_{2} \rightarrow C_{2}-C_{3}\right]\) 
Expanding along R₁ we get
\((a+b+c)[1\{(b-a)(a-b)-(c-a)(c-b)\}] \)
\( =(a+b+c)\left(b a-b^{2}-a^{2}+a b-c^{2}+c b+a c-a b\right) \)
\(=\frac{-1}{2}(a+b+c) \times(-2)\left(-a^{2}-b^{2}-c^{2}+a b+b c+c a\right) \) 
[multiply and divide by (-2)]
\(=\frac{-1}{2}(a+b+c)\left[a^{2}+b^{2}+c^{2}-2 a b-2 b c\right. \left.-2 c a+a^{2}+b^{2}+c^{2}\right] \)
\(=-\frac{1}{2}(a+b+c)\left[a^{2}+b^{2}-2 a b+b^{2}+c^{2}\right. \)
\(=\frac{-1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \) 
Also, \(\Delta=0\) 
\(\Rightarrow \frac{-1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0\)
\(\Rightarrow(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0[\because a+b+c \neq 0, \text { given }]\)
\(\Rightarrow a-b=b-c=c-a=0\)
\(\Rightarrow a=b=c \)

Given A = \(\left[ \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix} \right] \)
and B = \(\left[ \begin{matrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{matrix} \right] \)
\(\therefore AB=\left[ \begin{matrix} 2+4 & 2-2 & -4+4 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ -4+4 & 2-2 & -4+10 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix} \right] \)
\(\therefore \) AB = 6I
Premultiplying by A^-1
A^-1AB = 6A^-1I
=> IB = 6A^-1I  (\(\therefore \) A^-1A = I)
=> B = 6A^-1   (\(\therefore \) IX = X)
=> A^-1 = 1/6B
Given equations are:
x - y = 3
2x + 3y + 4z = 17
y + 2z = 7
=> \(\left[ \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 17 \\ 7 \end{matrix} \right] \)
AX = C, where
C = \(\left[ \begin{matrix} 3 \\ 17 \\ 7 \end{matrix} \right] \)
X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
=> A^-1AX = A^-1C
=> X = A^-1C
=> X = \(\frac { 1 }{ 6 } \left[ \begin{matrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 17 \\ 7 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} 12 \\ -6 \\ 24 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 4 \end{matrix} \right] \)
\(\therefore \) x = 2, y = -1 and z = 4

For intersection points, substitute
y² = 4x ⇒ y = 2√x
x² = 4y
x² = 4x 2√x
x⁴ = 64x
⇒ x(x³-64) = 0
⇒ x = 0
and x = 士 4
when x = 0, y² = 4 x 0⇒ y = 0
when x = 4, y² = 4 x 4 ⇒ y = 4
Points of intersection are (0, 0) and (4, 4).
Given, y² = 4x or y = 2√x = f(x)
and y = \(\frac{1}{4}x^{2}=g(x)\)

Let the company manufactures sweaters of type A = x, type B = y, daily.
\(\therefore\)  LPP is maximize. P = 200x + 20y s.t.
360x + 120y \(\le \) 72000 
\(\Rightarrow\) 3x + y \(\le \) 300
x + y \(\le \) 300
x - y  \(\le \) 100
\(3x+y=600,\begin{cases} x=0,y=600 \\ y=0,x=200 \end{cases}\)
\(x+y=300,\begin{cases} x=0,y=300 \\ y=0,x=300 \end{cases}\)
\(x-y=100,\begin{cases} x=100,\quad y=0 \\ y=100,\quad x=200 \end{cases}\)
\(\\ x\ge 0\)
\(y\ge 0\)

Getting vertices of feasible region as, O(0, 0), A(100, 0), B(175, 75), C(150, 150) and D(0, 300)
Maximum profit is P = 200(175) + 20(75)
= 35000 + 1500 = Rs. 36500

Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 }\quad and\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA\cos ^{ 2 }{ B } +cosB & cosC-cosA\cos ^{ 2 }{ C } -cosC \\ \cos ^{ 2 }{ A } +cosA & \cos ^{ 2 }{ A } -cosA & -\cos ^{ 2 }{ A } -cosA \end{matrix} \right| =0\)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA & cosC-cosA \\ \cos ^{ 2 }{ A } -1 & \cos ^{ 2 }{ B } -\cos ^{ 2 }{ A } & \cos ^{ 2 }{ C } -cosA \end{matrix} \right| =0\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }\)
(cos B - cos A ) x ( cos C -cos B)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & 1 & 0 \\ \cos ^{ 2 }{ A } -1 & cosB+cosA & cosC-cosA \end{matrix} \right| =0\)
\(\therefore \quad (cosB-cosA)\times (cosC-cosA)\times (cosC-cosB)\)
[1-0]=0  (Expanding along C₃)
\(\therefore \quad -cosB=cosA\quad or\quad cosC=cosAorcosC=cosB\)
\(\Rightarrow cosB=cosAorcosC=cosAorcosC=cosB\)
\(\Rightarrow \angle B=\angle Aor\angle C=\angle Aor\angle C=\angle B\)
\(\Rightarrow \triangle ABC\)  is an isosceles triangle.

Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 }\quad and\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA\cos ^{ 2 }{ B } +cosB & cosC-cosA\cos ^{ 2 }{ C } -cosC \\ \cos ^{ 2 }{ A } +cosA & \cos ^{ 2 }{ A } -cosA & -\cos ^{ 2 }{ A } -cosA \end{matrix} \right| =0\)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA & cosC-cosA \\ \cos ^{ 2 }{ A } -1 & \cos ^{ 2 }{ B } -\cos ^{ 2 }{ A } & \cos ^{ 2 }{ C } -cosA \end{matrix} \right| =0\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }\)
(cos B - cos A ) x ( cos C -cos B)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & 1 & 0 \\ \cos ^{ 2 }{ A } -1 & cosB+cosA & cosC-cosA \end{matrix} \right| =0\)
\(\therefore \quad (cosB-cosA)\times (cosC-cosA)\times (cosC-cosB)\)
[1-0]=0  (Expanding along C₃)
\(\therefore \ -cosB=cosA\quad or\quad cosC=cosAorcosC=cosB\)
\(\Rightarrow cosB=cosAorcosC=cosAorcosC=cosB\)
\(\Rightarrow \angle B=\angle Aor\angle C=\angle Aor\angle C=\angle B\)
\(\Rightarrow \triangle ABC\)  is an isosceles triangle.

Let the sides of rt. ΔABC be x and y.
\(\therefore\ x^2+y^2=4r^2\)
and A = area of \(\Delta=\frac{1}{2}xy\)
Let, \(S=A^2 =\frac{1}{4}x^2y^2\)
\(=\frac{1}{4}x^2(4r^2-x^2)\)
\(=\frac{1}{4}(4r^2x^2-x^4)\)
\(\therefore\ \frac{dS}{dx}=\frac{1}{4}[8r^2x-4x^3]\)
\(\Rightarrow \ \frac{dS}{dx}=0\Rightarrow x^2=2r^2 \ or\ x=\sqrt{2r}\)
\(y^2=4r^2-2r^2=2r^2\Rightarrow y=\sqrt2r\)
i.e. x = y and  \(\frac{d^2S}{dx^2}=(2r^2-3x^2)=2r^2-6r^2<0\)
⇒ Area is maximum when Δ is isosceles 

Given, volume of the box = 1024 cm³.Let length of the side ofsquare base be x cm and height ofthe box be y cm.
= x cm,
height of box = h cm.
Volume of box = 1024 cm³
\(\Rightarrow x.x.h=1024\)
\(\therefore h=\frac{1024}{x^2}\)
C (cost of box) = 5(2x²) + 2.5(4xh)
= 10x² + 10hx
\(=10x^2+\frac{10,240}{x}\)
\(\Rightarrow \frac{dC}{dx}=20x-\frac{10,240}{x^2}\)
Solving \(\frac{dC}{dx}=0,\)  we get x³ = 512   ஃ x = 8
\(=20+\frac{1(10240)}{x^3}>0\)
Thus, cost of box is least at x = 8 and least cost of
box is:
\(C(8)=10(8)^2+\frac{10240}{8}\)
= Rs. 1,920