Verified answer

Answer:

[tex]\qquad\qquad\qquad\boxed{ \bf{ \:(2) \: \: 500 \: \: }} \\ \\ [/tex]

Step-by-step explanation:

Given that,

[tex]\sf \: \sqrt{a} - \sqrt{b} = 20 \\ \\ [/tex]

[tex]\bf\implies \sqrt{a} = 20 + \sqrt{b} \\ \\ [/tex]

Now, Let assume that

[tex]\sf \: S = a - 5b \\ \\ [/tex]

can be rewritten as

[tex]\sf \: S = {( \sqrt{a} )}^{2} - 5b \\ \\ [/tex]

[tex]\sf \: S = {( 20 + \sqrt{b} )}^{2} - 5b \\ \\ [/tex]

[tex]\sf \: S = 400 + b + 40 \sqrt{b} - 5b \\ \\ [/tex]

[tex]\sf \: S = 400 + 40 \sqrt{b} - 4b \\ \\ [/tex]

On differentiating both sides w. r. t. b, we get

[tex]\sf \: \dfrac{dS}{db} = 0 + 40 \times \dfrac{1}{2 \sqrt{b} } - 4\\ \\ [/tex]

[tex]\sf \: \dfrac{dS}{db} = \dfrac{20}{\sqrt{b} } - 4\\ \\ [/tex]

For maxima or minima,

[tex]\sf \: \dfrac{dS}{db} = 0\\ \\ [/tex]

[tex]\sf \: \dfrac{20}{\sqrt{b} } - 4 = 0\\ \\ [/tex]

[tex]\sf \: \dfrac{20}{\sqrt{b} } = 4\\ \\ [/tex]

[tex]\sf \: \sqrt{b} = 5 \\ \\ [/tex]

[tex]\bf\implies \: \: b = 25 \\ \\ [/tex]

Now, we have

[tex]\sf \: \dfrac{dS}{db} = \dfrac{20}{\sqrt{b} } - 4\\ \\ [/tex]

[tex]\sf \: \dfrac{ {d}^{2} S}{d {b}^{2} } = \dfrac{20}{ - 2b\sqrt{b} } \\ \\ [/tex]

[tex]\sf \: \dfrac{ {d}^{2} S}{d {b}^{2} } = - \: \dfrac{10}{b\sqrt{b} } \\ \\ [/tex]

[tex]\sf \: \dfrac{ {d}^{2} S}{d {b}^{2} } \: at \: (b = 25) = - \: \dfrac{10}{25\sqrt{25} } \\ \\ [/tex]

[tex]\sf\implies \sf \: \dfrac{ {d}^{2} S}{d {b}^{2} } \: at \: (b = 25) < 0 \\ \\ [/tex]

[tex]\sf\implies \sf \:S \: is \: maximum \\ \\ [/tex]

Now, on substituting the value of b in

[tex]\sf \: \sqrt{a} = 20 + \sqrt{b} \\ \\ [/tex]

we get

[tex]\sf \: \sqrt{a} = 20 + \sqrt{25} \\ \\ [/tex]

[tex]\sf \: \sqrt{a} = 20 + 5 \\ \\ [/tex]

[tex]\sf \: \sqrt{a} = 25 \\ \\ [/tex]

[tex]\sf\implies \bf \: a= 625 \\ \\ [/tex]

So, Now Consider

[tex]\sf \: a - 5b \\ \\ [/tex]

[tex]\sf \: = \: 625 - 5 \times 25 \\ \\ [/tex]

[tex]\sf \: = \: 625 - 125\\ \\ [/tex]

[tex]\sf \: = \: 500\\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \: Maximum \: value \: of \: a - 5b = 500\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used

[tex]\sf \: \dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} } \\ \\ [/tex]

[tex]\sf \: \dfrac{d}{dx}\dfrac{1}{ \sqrt{x} } = - \dfrac{1}{2x \sqrt{x} } \\ \\ [/tex]

[tex]\sf \: \dfrac{d}{dx}k = 0\\ \\ [/tex]