The maximum value of $x y$ when $x+2 y=8$ is

The maximum value of $x y$ when $x+2 y=8$ is
  1. 20
  2. 16
  3. 24
  4. 8

Solution

$\begin{aligned} & x+2 y=8 \\ \therefore \quad & 2 y=8-x \\ \therefore \quad & y=\frac{8-x}{2} \end{aligned}$ Let $\mathrm{f}(x)=x y$ $\therefore \quad \mathrm{f}(x)=x \cdot \frac{(8-x)}{2}$ Differentiating w.r.t $x$, we get $\begin{aligned} \mathrm{f}^{\prime}(x) & =\frac{(8-x)-x}{2} \\ \mathrm{f}^{\prime}(x) & =4-x \end{aligned}$ To find critical points, $\begin{array}{ll} & \mathrm{f}^{\prime}(x)=0 \\ \therefore \quad & 4-x=0 \\ \therefore \quad & x=4 \end{array}$ critical point at $x=4$ $\therefore \quad \mathrm{f}(4)=\frac{4(8-4)}{2}=8$ $\therefore \quad$ Maximum value of the given function is 8 .

Asked in: MHT CET 2023 (09 May Shift 1)

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