What is the minimum value of #9sec^2x+16cos^2x#?

Question

Carter Anderson · Accepted Answer

The minimum value is #24#.

The problem is solved here using "completing the square". First note that we may render the given expression as: #9 sec^2 x+16cos^2 x = a^2+b^2# #a = 3 sec x = 3/cos x# #b = 4 cos x# Now recall the identity #(a\pm b)^2=a^2\pm 2ab+b^2# Select the minus sign and put in the values for #a# and #b# given above. Then #2ab# turns out to be a constant, #24#, and we have: #(3 sec x -4 cos x)^2=9 sec^2 x-24+16cos^2 x# #9 sec^2 x+16cos^2 x=color(blue)((3 sec x -4 cos x)^2)+24# The blue term, a squared quantity derived from the original expression, is called a completed square. We are sure that #9 sec^2 x+16cos^2 x# has to be at least #24# because the completed square is nonnegative. But is #24# the real minimum, or might the minimum be greater because the completed square doesn't get all the way down to zero? We find out by seeing whether there are real values of #x# where the completed square hits zero, thus: #3 sec x -4 cos x=0# #3/cos x = 4 cos x# #cos^2 x = 3/4# For real values of #x#, #cos^2 x# can take on any value between #0# and #1#. Thus we are sure that the completed square hits zero and the minimum of the given function is exactly #24#.