maximize q = xy, where x and y are positive numbers such that x + 25/3y² = 16. write the objective function…

Explanation:

Step1: Differentiate Q with respect to y

Given $Q = 16y-\frac{25}{3}y^{3}$, then $Q^\prime=\frac{dQ}{dy}=16 - 25y^{2}$

Step2: Set the derivative equal to zero

$16 - 25y^{2}=0$. Rearranging gives $25y^{2}=16$, so $y^{2}=\frac{16}{25}$. Since $y>0$, then $y = \frac{4}{5}$

Step3: Find the second - derivative of Q

$Q^{\prime\prime}=\frac{d^{2}Q}{dy^{2}}=- 50y$. When $y=\frac{4}{5}$, $Q^{\prime\prime}=-50\times\frac{4}{5}=-40<0$, which means Q has a maximum at $y = \frac{4}{5}$

Step4: Find the value of x

From $x+\frac{25}{3}y^{2}=16$, substitute $y = \frac{4}{5}$. Then $x+\frac{25}{3}\times(\frac{4}{5})^{2}=16$. So $x+\frac{25}{3}\times\frac{16}{25}=16$, $x + \frac{16}{3}=16$, and $x=\frac{32}{3}$

Step5: Calculate the maximum value of Q

$Q = xy$, substituting $x=\frac{32}{3}$ and $y=\frac{4}{5}$, we get $Q=\frac{32}{3}\times\frac{4}{5}=\frac{128}{15}$

Answer:

$\frac{128}{15}$