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Diagonal Matrix

If A = dia. d...

Question

If $A =$ dia. $(d_{1}, d_{2}, d_{3}, d_{4})$ where $d_{i} > 0 \forall i = 1, 2, 3, 4$ is a diagonal matrix of order $4$ such that $d_{1} + 2 d_{2} + 4 d_{3} + 8 d_{4} = 16,$ then the maximum value of $f (x) = {log}_{(tan x + cot x)} (det (A))$ where $x \in (0, \frac{π}{2})$ is equal to

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Solution

Given, $A =$ dia. $(d_{1}, d_{2}, d_{3}, d_{4})$
$\Rightarrow A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} d_{1} & 0 & 0 & 0 0 & d_{2} & 0 & 0 0 & 0 & d_{3} & 0 0 & 0 & 0 & d_{4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$
$∴ det (A) = d_{1} \cdot d_{2} \cdot d_{3} \cdot d_{4}$

Now, using A.M $\geq$ G.M. for positive numbers $d_{1}, 2 d_{2}, 4 d_{3}, 8 d_{4},$ we get
$\frac{d_{1} + 2 d_{2} + 4 d_{3} + 8 d_{4}}{4} \geq (d_{1} .2 d_{2} .4 d_{3} .8 d_{4})^{1 / 4}$
$\Rightarrow \frac{16}{4} \geq 2^{3 / 2} (d_{1} \cdot d_{2} \cdot d_{3} \cdot d_{4})^{1 / 4} \Rightarrow d_{1} . d_{2} . d_{3} . d_{4} \leq 4$
$\Rightarrow det (A) \in (0, 4]$

Now,
$f (x) = {log}_{(tan x + cot x)} (det (A)), x \in (0, \frac{π}{2})$
$= \frac{log (det (A))}{log (tan x + cot x)}$
As, $(tan x + cot x) \geq 2 \forall x \in (0, \frac{π}{2})$

$∴ f (x)$ will be maximum when numerator is maximum and denominator is minimum.

Hence, $f (x)_{max} = \frac{log 4}{log 2} = {log}_{2} (4) = 2$

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