Given that #7 cos x + 24 sin x = R cos (x - theta)# where R > 0 and theta is acute, how do you find the minimum and maximum values of #7 cos x + 24 sin x - 12# and the corresponding values of x?

1 Answer
Aug 4, 2018

#min.{7cosx+24sinx-12}=-37, and, #

#max.{7cosx+24sinx-12}=13#.

Explanation:

Given that, #7cosx+24sinx=Rcos(x-theta), R gt 0, theta" acute"#.

We expanding #cos(x-theta)#, & get,

#7cosx+24sinx=Rcosxcostheta+Rsinxsintheta#.

#"Comparing the respective co-efficients of "cosx and sinx,"#

we have, #Rcostheta=7 and Rsintheta=24#.

Squaring & adding , #R^2(cos^2theta+sin^2theta)=7^2+24^2, #

# or, R^2=25^2," giving, "R=+25...[because, R gt 0]#.

Now, #R=25rArrcostheta=7/R=7/25, &," similarly, "sintheta=24/25#.

Alternatively, #tantheta=24/7. :. theta=arctan(24/7)#.

Altogether, we have,

#7cosx+24sinx=25cos(x-theta), theta=arctan(24/7)#.

Knowing that, #-1 le cos(x-theta) le 1,# we, on multiplication

by #25 gt 0#, have,

# -25 le 25cos(x-theta) le 25, #

# i,e., -25 le 7cosx+24sinx le 25#.

Adding #-12, -37 le 7cosx+24sinx-12 le 13#.

Clearly, #min.{7cosx+24sinx-12}=-37, and, #

#max.{7cosx+24sinx-12}=13#.