Maximum and Minimum Values
• y =(1/4)x2 • x = 2y2 Next, we solve the system of equations. y =(1/4)(2y2)2 =) y = y4 If y =1,thenx = 2. If y =0,thenx =0. Thecriticalpoints (2,1) and (0,0). We now calculate the second derivatives to classify the critical point. • fxx(x,y)=6x • fyy(x,y)=48y • fxy(x,y)=12 138 of 155 ................
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