MHT CET · Maths · Application of Derivatives
The maximum value of \(x y\) when \(x+2 y=8\) is
- A 20
- B 16
- C 24
- D 8
Answer & Solution
Correct Answer
(D) 8
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& x+2 y=8 \\
\therefore \quad & 2 y=8-x \\
\therefore \quad & y=\frac{8-x}{2}
\end{aligned}\)
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 .
& 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 .
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