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  • Let a, b, c and d be positive real numbers such that .If the maximum value ofn

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} c^{2} d is 3750 \beta, then the value of \beta is 

Option: 1

55


Option: 2

108


Option: 3

90


Option: 4

110


Answers (1)

best_answer

Given \: \; \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=11 \quad$ (a, b, $c, \mathrm{~d}>0\}

\left(a^{5} \mathrm{~b}^{3} c^{2} d\right)max.= ?
Let assume Numbers –
\frac{\mathrm{a}}{5}, \frac{\mathrm{a}}{5}, \frac{\mathrm{a}}{5}, \frac{\mathrm{a}}{5}, \frac{\mathrm{a}}{5}, \frac{\mathrm{b}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{c}}{2}, \frac{\mathrm{c}}{2}

We know A.M. \geq G.M.
\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}+d}{11} \geq\left(\frac{a^{5} b^{3} c^{2} d}{5^{5} \cdot 3^{3} \cdot 2^{2} \cdot 1}\right)^{\frac{1}{11}}
\frac{11}{11} \geq\left(\frac{a^{5} b^{3} c^{2} d}{5^{5} \cdot 3^{3} \cdot 2^{2} \cdot 1}\right)^{\frac{1}{11}}

a^{5} \cdot b^{3} \cdot c^{2} \cdot d \leq 5^{5} \cdot 3^{3} \cdot 2^{2}

\max \left(\mathrm{a}^{5} \mathrm{~b}^{3} \mathrm{c}^{2} \mathrm{~d}\right)=5^{5} \cdot 3^{3} \cdot 2^{2}=337500

=90 \times 3750=\beta \times 3750
\beta=90

Option (C) 90 correct

Posted by

Deependra Verma

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