\(x+3 y=\pm 8\)

15

For equally inclined, consider \(\frac{d y}{d x}=\pm 1\)  (4,4)

Maximum slope m=-3+6+9=12

\(x \rightarrow \text { side, } h \rightarrow \text { hypotenuse }\)
\(x+h=G \text { (given) }\)
\(\text { Area of the right-angled triangle, }\)
\(A =\frac{1}{2} x \cdot \sqrt{h^{2}-x^{2}} \)
\(=\frac{1}{2} x \sqrt{(G-x)^{2}-x^{2}}=\frac{1}{2} x \sqrt{G^{2}-2 G x} \)
\(\text { If } A \text { is maximum, then } A^{2} \text { is maximum }\)
\(A^{2} =B(\text { say })=\frac{1}{4} x^{2}\left(G^{2}-2 G x\right) \)
\(=\frac{1}{4}\left(G^{2} x^{2}-2 G x^{3}\right) \)
\(\frac{d B}{d x} =\frac{1}{4}\left(2 x G^{2}-6 G x^{2}\right)\)
\(\text { For maximum area, } \frac{d B}{d x}=0\)
\(\Rightarrow 2 x G^{2}=6 G x^{2} \)
\(\Rightarrow G=3 x\)
\( \frac{d^{2} B}{d x^{2}}=\frac{1}{4}\left(2 G^{2}-12 G x\right) \)
\(\Rightarrow \left.\frac{d^{2} B}{d x^{2}}\right]_{G=3 x}<0 \)
\(\Rightarrow \text { area is maximum for } G=3 x\)
\(\Rightarrow x+h=3 x \Rightarrow h=2 x\)
\([\text { from }(i)]\Rightarrow \frac{x}{h}=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} \)

\(0.018\pi cm^3\)

a²-b²=0

We have \( f(x) =\tan ^{-1}(\sin x+\cos x), x>0 \)
\(f^{\prime}(x) =\frac{1}{1+(\sin x+\cos x)^2}(\cos x-\sin x)\)(on simplification)
\(=\frac{\cos x-\sin x}{2+\sin 2 x}\)
Note that 2 + sin 2x > 0 for all x in \(0, \frac{\pi}{4}\)
Therefore f ′(x) > 0 if cos x – sin x > 0
or f ′(x) > 0 if cos x > sin x or cot x > 1
Now \(\cot x>1 \text { if } \tan x<1 \text {, i.e., if } 0
Thus \(f^{\prime}(x)>0 \text { in }\left(0, \frac{\pi}{4}\right)\)
Hence f is increasing function in \(\left(0, \frac{\pi}{4}\right)\)

The area A₁ of square of side x is given by A₁= x²
and area A₂ of square of side y is given by
\(A_{2}=y^{2}=\left(x-x^{2}\right)^{2} \)
\(\frac{d A_{1}}{d x} =2 x, \frac{d A_{2}}{d x}=2\left(x-x^{2}\right)(1-2 x) \)
\(\frac{d A_{2}}{d A_{1}} =\frac{d A_{2}}{d x}= \frac{d A_{1}}{d x}=\frac{2\left(x-x^{2}\right)(1-2 x)}{2 x} \)
\(=(1-x)(1-2 x)=1-3 x+2 x^{2} \)

\(c^3\over6\sqrt{3}\) cu units