From the second equation we getx=-y-7λy4λ.Substituting the value ofxin the firstequation, we get-(1 +λ)(1 + 7λ)y2λ+ 8λy= 0 or 16λ2-(1 + 7λ)(1 +λ) = 0, sincey6= 0 —note thaty= 0⇒x= 0 which is not possible becausex2+ 8xy+y2= 225. Thus we get9λ2-8λ-1 = 0, soλ= 1,-19.We shall now show thatλ= 1 is not possible.λ= 1⇒4x+ 8y= 0⇒x=-2y⇒4y2-16y2+ 7y2= 225, which is not possible, thusλ6= 1.λ=-19⇒2x-29x-89y= 0⇒16x-8y= 0⇒y= 2x. Thusx2+ 16x2+ 28y2= 225⇒x=±√5, y=±2√5. Thus stationary points are (√5,2√5),(-√5,-2√5).F(x, y)=x2+y2-19(x2+ 8xy+ 7y2-225)=19h8x2+ 2y2-8xy+ 225i=19h2(2x-y)2+ 225iwhich is minimized when 2x=y. Thus the shortest distance is√25 = 5.Question 3(b)Show that the double integralZZRx-y(x+y)3dx dydoes not exist overR=[0,1 : 0,1].Solution.For a fixedx6= 0, the functionx-y(x+y)3is a bounded function ofyandZ10x-y(x+y)3dy=Z10-x+y(x+y)3+2x(x+y)3dy=1x+y+2x-2(x+y)210=11 +x-x(1 +x)2=1(1 +x)2NowZ10dx(1 +x)2= lim→0+Z1(1 +x)-2dx= lim→0+-12+11 +=12ThusZ10Z10x-y(x+y)3dy dx=12.Similarly, ify6= 0,Z10x-y(x+y)3dx=Z10x+y(x+y)3-2y(x+y)3dx=-1x+y-2y-2(x+y)210=-1(1 +y)25