Question

#QualityQuestion
@Trigonometry

Find the maximum value of 5cosA + 12sinA + 12.​

Answer 1

Answer:

Solution

5cosθ+12sinθ

52+122−−−−−−−√{552+122−−−−−−−√cosθ+1252+122−−−−−−−√sinθ}

sinϕ=552+122−−−−−−−√=513

cosϕ=1213=1252+122−−−−−−−√

putting it in the equation

13{sinϕcosθ+cosϕsonθ}

=13sin(θ+ϕ)

=12+13sin(θ+ϕ)

occurs when sin(θ+ϕ)=1

so,`12 + 13 xx1 = 121+ 13 = 25

Answer

Answer 2

Let the Function 5cosA + 12sinA + 12 = f(A).

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▪ Given :-

• f(A) = 5cosA + 12sinA + 12.

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▪ To Find :-

• Maximum Value of f(A).

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▪ Concept To Mind :-

• The maximum and minimum value of any function at the the point where First Derivative is Zero.

• for maximum value the sign of second derivative should be negative.

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▪ Solution :-

》We Have ,

$\large{f(A)} = 5 \cos A + 12 \sin A + 12$

$\bigstar \: \underline{ \pmb{ \mathfrak{ Differentiating \: \: both \: \: sides \: \: \text{w}.r.t \: \: \text{A} }}}$

$\large : \longmapsto f'(A) = - 5 \sin A + 12 \cos A$

$\bigstar \: \underline{ \pmb{ \mathfrak{ Differentiating \: \: both \: \: sides \: \: \text{w}.r.t \: \: \text{A} }}}$

$f''(A) = - 5 \cos A - 12 \sin A$

Now,

$\large \bigstar \: \underline{ \pmb{ \mathfrak{ For \: \: Maxima \: \: and \: \: Minima : }}} -$

$f'(A) = 0 \\ \\ : \longmapsto - 5 \sin A + 12\cos A = 0 \\ \\ : \longmapsto12 \cos A = 5 \sin A \\ \\ \large \pink{ : \longmapsto \boxed{\tan A = \frac{12}{5} }}\bf\:\:\:\:----(1)$

$: \longmapsto A = \tan {}^{ - 1} \bigg( \dfrac{12}{5} \bigg)$

$\large \bigstar \: \underline{ \pmb{ \mathfrak{ From \: \: (1) : }}} -$

$\tan A = \dfrac{12}{5} \\ \\ \sec {}^{2} A = 1 + \frac{144}{25} = \frac{169}{25} \\ \\ : \longmapsto \sec A = \frac{13}{5} \\ \\ : \longmapsto \boxed{ \cos A = \frac{5}{13}} \\ \\ : \longmapsto \sin A = \sqrt{1 - \frac{25}{169} } = \sqrt{ \frac{144}{169} } \\ \\ : \longmapsto \boxed{\sin A = \frac{12}{13} }$

Now ,

$f''( A) = - 5 \times \dfrac{5}{13} - 12 \times \frac{12}{13} \\ \\ = \bf - ve$

$\large\bigstar \: { \pmb{ \mathfrak{ \text{H}ence \: \: f( \text{A}) \: \: is \: \: maximum \: \: at }}} \\ \\ \large\bf A = \mathfrak{ { \tan}^{ - 1} \frac{12}{5} }$

So ,

$f( A)_{max} = 5 \times \frac{5}{13} + 12 \times \frac{12}{13} + 12 \\ \\ = \frac{25 + 144}{13} + 12 \\ \\ = \frac{169}{13} + 12 \\ \\ = 13 + 12 \\ \\ \Large\purple{ :\longmapsto  \underline {\boxed{{\bf f(A)_{maximum}=25} }}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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