Let a,b,c and d be positive real numbers such that a+b+c+d=11. If the maximum value of a^5 b^3 ^2d is 3750β, then the value of β is

Given:

$$a+b+c+d=11$$a+b+c+d=11

We know that,

$$\mathrm{AM}\geq \mathrm{GM}$$AM≥GM

$$\Rightarrow \frac{\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}+\frac{d}{1}}{11}\geq \sqrt[11]{\frac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}}$$⇒115a​+5a​+5a​+5a​+5a​+3b​+3b​+3b​+2c​+2c​+1d​​≥11553322a5b3c2d​​

$$\Rightarrow \frac{a+b+c+d}{11}\geq \sqrt[11]{\frac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}}$$⇒11a+b+c+d​≥11553322a5b3c2d​​

$$\Rightarrow \frac{11}{11}\geq \sqrt[11]{\frac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}}$$⇒1111​≥11553322a5b3c2d​​

$$\Rightarrow \sqrt[11]{\frac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}}\leq 1$$⇒11553322a5b3c2d​​≤1

$$\Rightarrow {a}^{5}{b}^{3}{c}^{2}d\leq {5}^{5}{3}^{3}{2}^{2}$$⇒a5b3c2d≤553322

$$\Rightarrow {a}^{5}{b}^{3}{c}^{2}d\leq 90\cdot 3750$$⇒a5b3c2d≤90⋅3750

So,

$$\beta =90$$β=90