Find the minimum value of 64 sec x + 27 csc x | vedclass

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Question

(C) Let $y = 64 \sec x + 27 \csc x$.To find the minimum value,differentiate $y$ with respect to $x$:$\frac{dy}{dx} = 64 \sec x 	an x - 27 \csc x \cot

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(C) Let $y = 64 \sec x + 27 \csc x$.To find the minimum value,differentiate $y$ with respect to $x$:$\frac{dy}{dx} = 64 \sec x 	an x - 27 \csc x \cot x$.Set $\frac{dy}{dx} = 0$ for critical points:$64 \sec x 	an x = 27 \csc x \cot x$.Substituting trigonometric identities:$64 \left(\frac{1}{\cos x}ight) \left(\frac{\sin x}{\cos x}ight) = 27 \left(\frac{1}{\sin x}ight) \left(\frac{\cos x}{\sin x}ight)$.$\frac{\sin^3 x}{\cos^3 x} = \frac{27}{64} \Rightarrow 	an^3 x = \left(\frac{3}{4}ight)^3$.Thus,$	an x = \frac{3}{4}$.For a right-angled triangle with opposite side $3$ and adjacent side $4$,the hypotenuse is $\sqrt{3^2 + 4^2} = 5$.Therefore,$\sec x = \frac{5}{4}$ and $\csc x = \frac{5}{3}$.Substituting these values into the original expression:$y = 64 \left(\frac{5}{4}ight) + 27 \left(\frac{5}{3}ight) = 16(5) + 9(5) = 80 + 45 = 125$.

Find the minimum value of $64 \sec x + 27 \csc x$ for $0 < x < \frac{\pi}{2}$.

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