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Find the maximum value of $$ 3^{\sin^2{\theta}} \cdot 27^{\cos^2{\theta}} + 8^{\sin{\theta}}\cdot 16^{\cos{\theta}} $$

Where does maximum of the expression occur?

I can find maxima of individual terms easily, but since they occur at different values of $\theta$, that is not getting anywhere.

If I differentiate, it becomes very tedious.

Is there any clever rearrangement or logic, without calculus, which will get me max value of this expression?

I tried arithmetic mean greater than geometric mean but could not get anywhere.

Thank you.

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You can rewrite the expression as $$y=3^{\cos (2 t)+2}+2^{3 \sin (t)+4 \cos (t)}$$

The first piece will reach a maximum value $(3^3=27)$ when $t=0$ and the second will reach a maximum value $(2^5=32)$ when $t=2 \cot ^{-1}(3)$. This means that the maximum value of $y$ will not exceed $27+32=59$; on the other side, the maximum value of $y$ will be larger than $3^3+2^4=43$.

Plotting the function, you could see that the maximum occurs "not very far" of $t=0$. Using composition of Taylor series (this is quite tedious), $$y=43+48 \log (2)\,t+ \left(72 \log ^2(2)-32 \log (2)-54 \log (3)\right)t^2+ O\left(t^3\right)$$ Computing the derivative and making it equal to zero gives $$t_{max}=-\frac{12 \log (2)}{36 \log ^2(2)-16 \log (2)-27 \log (3)}\approx 0.354603$$ which gives $y_{max}\approx 48.4354$.

A numerical optimization would lead to $t_{max}\approx 0.330073$ and $y_{max}\approx 48.4634$.

For sure, you could use Newton method to find the zero of $$y'=\log (2) 2^{3 \sin (t)+4 \cos (t)} (3 \cos (t)-4 \sin (t))-2 \log (3) \sin (2 t) 3^{\cos (2 t)+2}$$ and the iterates would be $$\left( \begin{array}{cc} n & t_n \\ 0 & 0.000000 \\ 1 & 0.354603 \\ 2 & 0.329806 \\ 3 & 0.330073 \end{array} \right)$$

Finally, being very lazy, using a small programmable pocket calculator set in degrees, using steps of $1 ^{\circ}$, the maximum of $y$ was reached for $t=19 ^{\circ}=0.331613$ radians and found to be $48.4633$.

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Write your term as $$3^{3\cos^2(\theta)+\sin^2(\theta)}+2^{3\sin(\theta)+4\cos(\theta)}$$

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  • $\begingroup$ that is how i can max of individual terms , then what ? $\endgroup$ – George Carlin Aug 14 at 7:15
  • $\begingroup$ I would use a substitution, for instance $$a=\sin(\theta),b=\cos(\theta)$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 14 at 9:55

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