Question

what is the maximum value of 3sina + 4cosa ?

Answer 1

Answer:

Therefore the maximum value of the function is 5.

explanation

The first derivative of the function will be:

f′(x)=4cosA−3sinA

f′(x) will be zero when 4cosA = 3sinA

or,43=tanAor,tanA=43=pbThen,sinA=45andcosA=35f(x)=4sinA+3cosA=4∗45+3∗35=16+95=5

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Answer 2

for maxima ,

we apply formula

$\sqrt{ {a}^{2} + {b}^{2} }$

so ,

max =

$\sqrt{ {3}^{2} + {4}^{2} } \\ \\ = \sqrt{9 + 16} \\ \\ = \sqrt{25} \\ \\ = 5$