Dabar pagal pitagora galim suzinot H = ([tex]\frac{16 - x^2}{2x})^2[/tex] - [tex](\frac{1}{2}x)^2 [/tex]= [tex]\frac{x^4 - 32x^2 + 256}{4x^2}[/tex] - [tex]\frac{1}{4}x^2[/tex] = [tex]\frac{x^4 - 32x^2 + 256 - x^4}{4x^2}[/tex] = [tex]\frac{16(16 - 2x^2)}{4x^2} = [/tex] 4trumpinas ir lieka [tex]\frac{4(16 - 2x^2)}{x^2} =[/tex] is po saknies tai bus toks rezultatas koks mum reikalingas>
2. [tex]V = \frac{1}{3}\cdot x^2 H[/tex], kur H tas rezultatas kuri mes gavom 3. Isvestine min max.