by what rational number should - 9 upon 35 be multiplied to get 3 upon 5
Answers
Step-by-step explanation:
------------------------------
▪ Given :-
f(A) = 5cosA + 12sinA + 12.
-------------------------------
▪ To Find :-
Maximum Value of f(A).
-------------------------------
▪ Concept To Mind :-
The maximum and minimum value of any function at the the point where First Derivative is Zero.
For Maxima the sign of second derivative should be negative.
-------------------------------
▪ Solution :-
》We Have ,
\large{f(A)} = 5 \cos A + 12 \sin A + 12f(A)=5cosA+12sinA+12
\bigstar \: \underline{ \pmb{ \mathfrak{ Differentiating \: \: both \: \: sides \: \: \text{w}.r.t \: \: \text{A} }}}★
Differentiatingbothsidesw.r.tA
Differentiatingbothsidesw.r.tA
\large : \longmapsto f'(A) = - 5 \sin A + 12 \cos A:⟼f
′
(A)=−5sinA+12cosA
\bigstar \: \underline{ \pmb{ \mathfrak{ Differentiating \: \: both \: \: sides \: \: \text{w}.r.t \: \: \text{A} }}}★
Differentiatingbothsidesw.r.tA
Differentiatingbothsidesw.r.tA
f''(A) = - 5 \cos A - 12 \sin Af
′′
(A)=−5cosA−12sinA
Now,
\large \bigstar \: \underline{ \pmb{ \mathfrak{ For \: \: Maxima \: \: and \: \: Minima : }}} -★
ForMaximaandMinima:
ForMaximaandMinima:
−
\begin{gathered}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)\end{gathered}
f
′
(A)=0
:⟼−5sinA+12cosA=0
:⟼12cosA=5sinA
:⟼
tanA=
5
12
−−−−(1)
: \longmapsto A = \tan {}^{ - 1} \bigg( \dfrac{12}{5} \bigg):⟼A=tan
−1
(
5
12
)
\large \bigstar \: \underline{ \pmb{ \mathfrak{ From \: \: (1) : }}} -★
From(1):
From(1):
−
\begin{gathered} \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} }\end{gathered}
tanA=
5
12
sec
2
A=1+
25
144
=
25
169
:⟼secA=
5
13
:⟼
cosA=
13
5
:⟼sinA=
1−
169
25
=
169
144
:⟼
sinA=
13
12