Respuesta :
Given the equation:
[tex]y=x^2+\frac{54}{x}[/tex]Let's find the global maxima and minima values for the function on the interval [1, 108]
To find the global maxima on the interval, let's first find the derivative:
[tex]y^{\prime}=2x-\frac{54}{x^2}[/tex]Set the derivative to zero and solve for x:
[tex]\begin{gathered} 2x-\frac{54}{x^2}=0 \\ \\ \frac{2x\left(x^2\right)-54}{x^2}=0 \\ \\ 2x^3-54=0 \end{gathered}[/tex]Add 54 to both sides:
[tex]\begin{gathered} 2x^3-54+54=0+54 \\ \\ 2x^3=54 \\ \\ x^3=\frac{54}{2} \\ \\ x^3=27 \\ \\ Take\text{ the cuberoot of both sides:} \\ \sqrt[3]{x^3}=\sqrt[3]{27} \\ \\ x=3 \end{gathered}[/tex]Substitute 3 for x in the original function and solve for y:
[tex]\begin{gathered} y=x^2+\frac{54}{x} \\ \\ y=3^2+\frac{54}{3} \\ \\ y=9+18 \\ \\ y=27 \end{gathered}[/tex]Therefore, the global minima is: (3, 27)
To find the global maxima, solve for the following:
at x = 1 and at x = 108
[tex]\begin{gathered} y=1^2+\frac{54}{1}=54 \\ \\ \\ y=108^2+\frac{54}{108}=11664+0.5=11664.5 \\ \end{gathered}[/tex]Therefore, the global maxima is: (108, 11664.5)
ANSWER:
Global maximum: (108, 11664.5)
Global minimum: (3, 27)
