Question

find maxima or minima if y=sinx+cosx​

Answer 1

Explanation:

a*sinx + b*cosx, maximum & minimum value is (+,-)√[a^2+b^2]. As here a=b=1, maximum and minimum value of sinx + cosx = (+,-)√[1^2+1^2], Maximum value of this function is sqrt(2) and minimum is -sqrt(2).

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Answer 2

let y=cosx(sinx+cosx)

y=cosx×sinx+(cosx)^2

differentiate y with respect to x

dy/dx=(cosx)×(cosx)+(sinx)×(-cosx)+2×(cosx)×(-sinx)

dy/dx=(cosx)^2–3×(sinx)×(cosx)-(1)

dy/dx=cosx(cosx-3sinx)

for Maxima or minima dy/dx=0

So either cosx=0 or tanx=1/3

Differentiating (1) again

d^(2)y/dx^(2)=2×(cosx)×(-sinx)-3×[sinx×(-sinx)+cosx×(cosx)]-(2)

Putting cosx=0 and sinx=1 in above equation (2)

d^(2)y/dx^(2)=0–3×[(-1)+0]=3>0 hence minima occurs at cosx=0

Putting tanx=1/3 i.e sinx=1/sqrt(10) and cosx=3/sqrt(10) in equation (2)

d^2y/dx^2=2×(3/sqrt(10))×(-1/sqrt(10)-3×[-1/10+9/10]

d^(2)y/dx^(2)=(-6)/10+(-24)/10=-3<0 hence Maxima occurs at tanx=1/3

Putting this in y

so max value of y =3/sqrt(10)×(3/sqrt(10)+1/sqrt(10))=12/10=1.2.

Hope it helps..