If #M# and #m# be the maximum and minimum value of #f (x,y) = 7x^2+4xy +3y^2# subjected to the condition #x^2+y^2 =1# Find #[(M+m)/2]#?

1 Answer
Mar 15, 2017

#(m+M)/2=5#

Explanation:

Changing to polar coordinates using the pass transformations

#{(x=r costheta),(y=rsin theta):}#

we have

#f(x,y)->f(r,theta)=r^2(7cos^2theta+4sintheta cos theta+3sin^2theta)#

and the condition

#x^2+y^2=1->r^2=1# then the transformed problem is

#{min,max}f(1,theta) = {min,max}(7cos^2theta+4sintheta cos theta+3sin^2theta)#

but after simplifications

#7cos^2theta+4sintheta cos theta+3sin^2theta=5+2(sin(2theta)+cos(2theta))#

so

#f(1,theta)=5+2(sin(2theta)+cos(2theta))#

The stationary points condition is

#(df)/(d theta) = 4(sin(2theta)+cos(2theta))=0# and this is attained at

#theta = 1/2(pm pi/4 + 2kpi)# for #k=0,1,2,cdots#

The stationary points qualification is done evaluating

#(d^2f)/(d theta^2) = -8(sin(2theta)+cos(2theta))# resulting in

#{min,max}={8 sqrt[2], -8 sqrt[2]}#

The values at those points #theta = {-pi/8, pi/8}# are

#m = 5 - 2 sqrt[2]# and #M = 5 + 2 sqrt[2]# so

#(m+M)/2=5#