Problem

Maximize $Q=x^{2} y$, where $x+2 y=50$.

Answer

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Answer

Final Answer: The maximum value of the function \(Q=x^{2} y\) subject to the constraint \(x+2 y=50\) is \(\boxed{\frac{250000}{27}}\).

Steps

Step 1 :We are given the function \(Q=x^{2} y\) and the constraint \(x+2 y=50\).

Step 2 :We can express \(y\) from the constraint as \(y = 25 - \frac{x}{2}\).

Step 3 :Substitute \(y\) into the function \(Q\) to get \(Q = x^{2}(25 - \frac{x}{2})\).

Step 4 :Find the derivative of \(Q\) with respect to \(x\) to get \(Q' = -\frac{x^{2}}{2} + 2x(25 - \frac{x}{2})\).

Step 5 :Set the derivative equal to zero to find the critical points, which are \(x = 0\) and \(x = \frac{100}{3}\).

Step 6 :Substitute these critical points back into the function \(Q\) to find the maximum value, which is \(\frac{250000}{27}\).

Step 7 :Final Answer: The maximum value of the function \(Q=x^{2} y\) subject to the constraint \(x+2 y=50\) is \(\boxed{\frac{250000}{27}}\).

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