Find that maximum!

Let it be $8=R \sin a$ $15=R \cos a,$ so that, the original expression becomes $15 \sin x + 8 \cos x = R \left( \sin x \cos a + \sin a \cos x \right),$ which in turn, leads to $= R \sin\left( x+a\right).$ Solving the first system for $R$ by squaring both equations and adding them, we get $8^2 + 15^2 = R^2 \sin^2 a + R^2 \cos^2 a = R^2,$ because $\sin^2\alpha + \cos^2\alpha =1$ for any $\alpha$ . Hence, $R=17$ . The original expression is equivalent to $17\sin\left(x+a\right)$ but since $\left| \sin(x+a)\right|\leq 1$ , we find that the maximum it's exactly 17 .

$y=15sinx+8\cos x$ $\frac{ d }{ dx }(15sinx+8\cos x)=0$ $y'=15\cos x-8 \sin x=0$ $\tan x=\frac{ 15 }{ 8 }$ x is in the first quadrant as both sine and cosine functions are positive in 1st quadrant so we have, $\sin x=\frac{ 15 }{ 17 }$ $\cos x=\frac{ 8 }{ 17 }$ so $\frac{ 225 }{ 17 }+\frac{ 64 }{ 17 }=17$

15 sin x + 8 cos x ~~~~~~~~>Maximum
So by differentiating and putting the result equal to zero we get the value of x for maximum result (Maxima);
15 cos x - 8 sin x = 0
15 cos x = 8 sin x
sin x / cos x = 15/8
tan x = 15/8 ~~~~~~~~~~~~~~> x =61.92751306
Substituting in the equation we get; 15 sin 61.92751306 + 8 cos 61.92751306 = 17

We can solve this problem using trigonometric identity. First, guide the 15 sin x + 8 cos x * to the trigometric identity of * cos( a - b ) , that proved by same with:

Rcos( a - b ) = *Rcos a .cos b* + Rsin a .sin b **, assuming that R is an integer.

Now, compare this function at the problem with that trigonometric identity, and We can catch the pattern, delcare that x = Rcos a * = 8 and * y = Rsin a * = 15. Then, * x^{2} + y^{2} = R^{2}.((sin a )^{2} + (cos a )^{2}) . Since, (sin a )^{2} + (cos a )^{2} = 1 .

Hence, x^{2} + y^{2} = R^{2} as well as the maximum values of that function. So, the maximum value of that function is \sqrt{8^{2} + \sqrt{15^{2}} = 17 .

√(15^2+8^2)=√17^2=17

Solution by setting derivative = 0

y = 15sin(x) + 8cos(x)

y' = 15cos(x) - 8sin(x) = 0 --> tan(x) = $\frac{15}{8}$

Draw the triangle that represents tan(x) = $\frac{15}{8}$ , that is a right triangle with an angle x and the opposite side 15 and adjacent side 8. By Pythagorean Theorem, $hypotenuse^{2}$ = $15^{2}$ + $8^{2}$ --> hypotenuse = 17. Now evaluate y using this triangle y = 15sin(x) + 8cos(x) = 15( $\frac{15}{17}$ ) + 8( $\frac{8}{17}$ ) = 17

sqrt{(15)^2+8^2}=17

Divide and Multiply the expression with 17..

or, the given expression becomes 17[ 15/17(sinx )+ 8/17(cosx)]

let 15/17 = siny

then 8/17= cosy

Puting the values in the above expression we get 17[siny.sinx+ cosy.cosx] = 17[cos(x-y)]

Max. value of cos(x-y)=1

Therefore the max value of the given expression is 17.

Graphing the function, the largest value is y= 17 when x = 1.08, very close to pi/3.

Using calculus, you can find the derivative , but with algebra, graphing is the best way.

Y' = 15cosx -8sinx = 0 15cosx = 8 sinx 15/8 = tanx X= tan^-1(15/8) X= 1.0808... Then f(1.0808) = 17

First off, by finding the first derivative of the expression we have

$y'=15cosx-8sinx \Rightarrow 15cosx=8sinx \Rightarrow tgx=15/8$

Therefore, since $arctgx=1,08$ we can come to a conclusion that $x=61,92$ .

Furthermore, $f(61,92)=17$ which is also the maximum value of the equation.

For the expressions like +(or)-asinx+(or)-bcosx the maximum value is square root (a^2+b^2)

Use cauchy schwarz,the two sequences would be sinx,cosx and 15,8

a( sin x ) + b( cos x ) +c maximum c + [ a^2 + b^2 ]^(1/2)

This is a property which i learnt in class 11..the property is as such {-sqrt(a^2 + b^2)<=asin(x) + bcos(x)<=sqrt(a^2 + b^2)}.....this suggests the range of the asin(x) + bcos(x)...{where a and b are the coefficients of sin x and cos x respectively)

Graphing the function, the largest value is y= 17 when x = 1.08, very close to pi/3.

Using calculus, you can find the derivative , but with algebra, graphing is the best way.

Y' = 15 cos x -8 sin x = 0 15 cos x = 8 sin x 15/8 = tan x X= tan^-1(15/8) X= 1.0808... Then f(1.0808) = 17

$15\sin(x) +8\cos(x)$ can be written as $17(\frac{15}{17}\sin(x) +\frac{8}{15}\cos(x))$ . Let $\alpha$ be an angle such that $\tan(\alpha) = \frac{8}{15}$ , then the expression equals $17(\cos(\alpha)\sin(x)+\sin(\alpha)\cos(x))$ which is equivalent to $17\sin(\arctan(\frac{8}{15})+x)$ . Since $-1 \leq \sin(\theta) \leq 1$ , $-17 \leq 17\sin(\arctan(\frac{8}{15})+x) \leq 17$ . So the maximum value of the expression is $17$ .

$f(x)=15sinx+8cosx$

=> $f'(x)=15cos(x)-8sin(x)$

$f'(x)=0 => tan(x)=\frac{15}{8}$

Apply pythag $tan(x)=\frac{opposite}{adjacent} =>hypotenuse=(15^{2}+8^{2})^{\frac{1}{2}}=17$

So, at maximum point, $sin(x) = \frac{15}{17} ,cos(x)=\frac{8}{17}$

=> $f(x)=\frac{15\times15}{17}+\frac{8\times8}{17}=17.$

For:- [a * $\sin \theta$ ] + [b * $\cos \theta$ ]

Maximum Value: sqrt[ (b^2) + (a^2) ]

Minimum Value: sqrt[ (b^2) - (a^2) ]

for any question of the form a sin(x) + b cos (x) answer will be (a^2 + b^2)^(1/2) so according to question 15^2 = 225 8^2 = 64 225 + 64 = 289 square root of 289 will be 17 so 17 is answer

I'm plot in Graphmática!

But, the solution is

I'm derive (differenciacion) y=15sinx+8cosx y'=15cosx-8sinx

The maximo is 15cox-8sinx=0 15cosx=8sqrt(1-(cosx)^2)

Solve the equation in Maxima x=17

We can use a cos x + b sin x with (a^{2} +b^{2})^{1/2} * sin(X+ f) Then the maximum value if sin(x+f) is 1. So (25^{2} + 8^{2})^{1/2} = 17

Maximum Value of asinx+bcosx=Sqrt(a^2+b^2) hence 17

EquAl=17

We have $15\sin x+8\cos x=17\left( \frac{15}{17}\sin x+\frac{8}{17}\cos x\right)=17(\sin(\theta+x))$ where $\theta=\cos^{-1}\frac{15}{17}$ . We know that $17\sin(\theta+x)\le 17$ therefore the maximum value is $\boxed{17}$ .

Graphing the function, the largest value is y= 17 when x = 1.08, very close to pi/3.

Using calculus, you can find the derivative , but with algebra, graphing is the best way.

Y' = 15cosx -8sinx = 0 15cosx = 8 sinx 15/8 = tanx X= tan^-1(15/8) X= 1.0808... Then f(1.0808) = 17

Max value of a $\sin \theta$ +b $\cos \theta$ is [(a^2)+(b^2)]^0.5

Max value of a $\sin \theta$ +b $\cos \theta$ is [(a^2)-(b^2)]^0.5

To find the maximum value, we must differentiate it. d(15 sin x + 8 cos x) / dx = 0 ==> 15 cos x - 8 sin x = 0. Hence we get that tan x = 15/8. The value of x must be x = arc tan 15/8 = 61.92. Substitute this value : 15 sin 61.92 + 8 cos 61.92 = 16.99 = 17

Using Lagrange Notation for differentiation,

f(x) = 15sin(x) + 8cos(x) f'(x) = 15cos(x) - 8sin(x)

Let f'(x) = 0 to maximise the function so, 15cos(x) = 8sin(x) Hence, tan(x) = 15/8 Therefore, x = tan-1(15/8) so, 15sin(x) + 8cos(x) = 17

y= 15sinx + 8cosx

Take derivative of y to find the maxima, dy/dx = 15cosx- 8sinx =0

tanx = 15/8 ( for maximum y)

for a triangle with sides , 15, 8 and the third side 17 ( by pythagoras theorem)

substitute the values in original equation to get the maximum value of y

The maximum and minimum value of asinx + bcosx is (plus minus ) root(a^2+b^2) so root of (225+64) is 17

Graphing the function, the largest value is y= 17 when x = 1.08, very close to pi/3.

Using calculus, you can find the derivative , but with algebra, graphing is the best way.

Y' = 15cosx -8sinx = 0 15cosx = 8 sinx 15/8 = tanx X= tan^-1(15/8) X= 1.0808... Then f(1.0808) = 17

Let, f(x)=15sinx+8cosx .Then f '(x)=15cosx-8sinx.Now, for largest possible value f '(x)=0 then x=61.927513 . so f(x)=17.

15sinx+8cosx = 17 sin( x + p) where sinp =8/17 and cosp = 15/17 then the maximun values of sin(x+p) = 1 then the maximun values of 15sinx+8cosx will be 17

We know that the max. Value of the expression asinx + bcosx is square root of (a^2+b^2) comparing 15sinx + 8cosx to asinx +bcosx we get a=15 and b=8 so the max. Value is (15^2+8^2)^1/2 = 17

By observation as we can see that 15 and 8 are the parts of the pythagorean triplet (8,15,17).

8^2 + 15^2 = 17^2.

17(15/17 sinx + 8/17 cosx) ........(i)

Let sinA = 8/17. (A be acute angle.) Therefore, cosA = 15/17.

Expression (i) can be written as 17 sin(x + A). .........(ii)

And we know that value of sin(x + A) ranges from [0,1].

Therefore, maximum value of Expression (ii) is 17.

by differentiating the given equation we get tanx=15/8 from there sinx=15/17,cosx=8/17 so substitute these values in the given equation

the double derivative is negative is negative thus a maxima is to be obtained in a single derivative. the derivative is tanx=15/8 or sinx is 15/17 and cosx =8/17 thus putting in the values we get 17

Y' = 15cosx -8sinx = 0 15cosx = 8 sinx 15/8 = tanx X= tan^-1(15/8) X= 1.0808... Then f(1.0808) = 17

By making use of the R-formula,

15 sin x + 8 cos x = R sin (x + a) where R is a positive real number and a is an acute angle.

By expanding the RHS, one will get R sin x cos a + R cos x sin a

Equating the coefficients of sin x and cos x, 15 = R cos a and 8 = R sin a

By making use of the trigonometric identity sin^2 x + cos^2 x = 1, we have R^2 = 15^2 + 8^2 So, R = 17.

This means that we can rewrite the expression above as 15 sin x + 8 cos x = 17 sin (x + a)

It will be clear that 17 will be the largest possible value of the expression.

d(15sinx + 8cosx)/dx = 15cosx - 8sinx Set 15cosx - 8sinx = 0 Thus, 15cosx = 8sinx (x =/= n*pi/2) tan(x) = 15/8, x = arctan(15/8) Input the positive value of x

Graphing the function, the largest value is y= 17 when x = 1.08, very close to pi/3.

Using calculus, you can find the derivative , but with algebra, graphing is the best way.

Y' = 15cosx -8sinx = 0 15cosx = 8 sinx 15/8 = tanx X= tan^-1(15/8) X= 1.0808... Then f(1.0808) = 17

Y' = 15cosx -8sinx = 0 15cosx = 8 sinx 15/8 = tanx X= tan^-1(15/8) X= 1.0808... Then f(1.0808) = 17

Do it using Maxima and Minima

Let f(x) = 15 sin(x) + 8 cos(x)

d(f(x))/dx = 15cos - 8 sin(x)

for maxima d(f(x)) = 0

So 15sin(x) = 8cos(x)

so we get

tan(x) = (15/8)

Hence , sin(x) = (15/17) cos(x) = (8/17)

putting these values we get

maximum value of the given expression as 17

for max of acosx+bsinx we have y=root of ( a square + b square )

first method. for max. value differentiation of 15sinx+8cosx should be equal to 0. so d/dx(15sinx+8cosx)=0 solving this we have 15cosx-8sinx=o = tanx=15/8 so by this we have sinx=15/17 cosx=8/17 putting these values in
15sinx+8cosx we get the ans.i.e. 17.....

second method. we know that the max value of asinx + bcosx is √a^2+b^2 so comparing asinx + bcosx to 15sinx + 8 cosx we have a=15 b=8 so max. value is √15^2+8^2=√289 =17

15cosx -8sinx = 0 15cosx = 8 sinx 15/8 = tanx X= tan^-1(15/8) X= 1.0808... Then f(1.0808) = 17

f(x) = 15sin x + 8cos x f'(x) = 0 give the solution

For any general equation of form
$asinX + bcosX$ = $\sqrt( a^2+b^2)$ $(\frac{asinX}{\sqrt( a^2+b^2)} + \frac{bcosX}{\sqrt( a^2+b^2)})$

as a,b, $\sqrt( a^2+b^2)$ forms a pythagorean triplet $= \sqrt( a^2+b^2) ( cosKsinX+ sinKcosX)$

$=\sqrt( a^2+b^2)(sin(K+X)$ as sin t, t=K+X can have max value of 1 , therefore maxval of given equation $=\sqrt( a^2+b^2)$ On substituting a and b we get $\sqrt( a^2+b^2)$ = 17

A = 17 (sinasinx + cosacosx) = 17 cos(x-a) <= 17 with sina = 15/17

since the coefficient of sinx is the largest .try x=60 degree. else (when coefficient of cos is the largest) try x=30 degree.

Let 15=r sina and 8=r cosa,squaring and adding, 225+64=r^{2}sin^{2}a+r^{2}cos^{2}a =>r^{2}(sin^{2}+cos^{2})=289 =>r=17 So now,15sinx+8cosx=rsinasinx+rcosacosx=r(cosacosx+sinasinx)=r(cos(a-x))

Now cos x is max when it is 1.and r =17,so max value is 17*1=17

taking eq
15sinx+8cosx now divide it by underroot(15^2+8^2) we get 17(sinx 15/17 + cosx 8/17) it can be seen as 17(sinx cosy + cosx siny)

as siny^2 + cosy^2 =1 thus we get 17(sin(x+y)) we know that maximum value of sinq is 1 . taking sin(x+y) as 1 . we get 17 as answer.

In order to solve the problem we can consider the function f(x)=15sinx+8cosx . Since the function is differentiable (and thus continuous) for all values of x it´s largest value must have a derivative equal to zero. We take the first derivative of the function: f´(x)= 15cosx - 8sinx and make it equal to zero. We solve for x (take the smaller value)---> x= arctg (15/8). Use the second derivative to prove it is a maximun. The value of the function for this is 17.

Defina por f(x) = 15 sen x + 8 cos x , veja que queremos encontrar o máximo de f(x). Um truque muito utilizado quando se tem uma expressão desse tipo é o truque do triângulo retângulo, que trata-se da construção de um triângulo retângulo com catetos 15 e 8. Assim, a hipotenusa desse triângulo vale 17, logo temos as seguintes relações para um ângulo y qualquer desse triângulo: (I) 15 = 17cos y (II) 8 = 17sen y

Substituindo (I) e (II) em f(x) : f(x) = 17sen x.cos y + 17sen y.cos x f(x) = 17sen (x+y) 17 é uma constante, então o máximo de f(x) se dará no máximo de sen (x + y), que é facilmente encontrado, valendo 1. Logo, máx(f(x)) ; 17.1 = 17

we know that $\tan \theta$ = \frac{\(\sin \theta }{ $\cos \theta$ })

by knowing this, we can use the Pythagorean Theorem: \sqrt{\(15^{2} + $8^{2}$ })

which is equal to 17 ^_^

acosx+bsinx=sqrt(a^2+b^2)*cos(x-p)

8cosx+15sinx=sqrt(8^2+15^2) cos(x-p)=17cos(x-p) since maximum of cos(x-p)=1 then the maximum 17cos(x-p)=17 1=17

sin45=cos45 & it is the largest value.so x should be 45