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Find the smallest value of a constant making a function meet given conditions

Consider the function

    \[ f(x) = 5x^2 + Ax^{-5} \qquad A > 0. \]

Find the minimum value of A such that f(x) \geq 24 for all x > 0.


For this problem, first we want to find where the function has a minimum. Then, we’ll set this minimum equal to 24 to solve the problem.

To find the minimum we take the derivative of f(x),

    \[ f'(x) = 10x - 5Ax^{-6}. \]

Setting this equal to 0 we have

    \[ 10x - 5Ax^{-6} = 0 \quad \implies \quad 10x^7 = 5A \quad \implies \quad x = \left( \frac{A}{2} \right)^{\frac{1}{7}}. \]

So, f(x) has a minimum at this value of x. Now we plug this value of x into f(x) and set it equal to 24 (so that f(x) = 24 at its minimum).

    \begin{align*}  &&5 \left( \frac{A}{2} \right)^{\frac{2}{7}} + A \left( \frac{A}{2} \right)^{-\frac{5}{7}} &= 24 \\ \implies && \left( \frac{A}{2} \right)^{\frac{2}{7}} \left( 5 + A \left( \frac{A}{2} \right)^{-1} \right) &= 24 \\ \implies && \left( \frac{A}{2} \right)^{\frac{2}{7}} (7) &= 24 \\ \implies && \left( \frac{A}{2} \right)^{\frac{2}{7}} &= \frac{24}{7} \\ \implies && \frac{A}{2} &= \left( \frac{24}{7} \right)^{\frac{7}{2}} \\ \implies && A &= 2  \left( \frac{24}{7} \right)^{\frac{7}{2}}. \end{align*}

Thus, f(x) \geq 24 for all x > 0.

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