Question

If x and y are both non-negative and if x+y=π, then the maximum value of 5sinxsiny is equal to

• A. 1
• B. √5
• C. 5
• D. -5
• E. 0

Answer 1

The Correct Option is C

We are given the equation: \[ 5 \sin x \sin y \] and the condition that \( x + y = \pi \).
Step 1: Use the trigonometric identity We can use the following trigonometric identity to simplify the expression for \( \sin x \sin y \): \[ \sin x \sin y = \frac{1}{2} [\cos(x - y) - \cos(x + y)]. \] Since \( x + y = \pi \), we know that \( \cos(x + y) = \cos(\pi) = -1 \). Therefore: \[ \sin x \sin y = \frac{1}{2} [\cos(x - y) - (-1)] = \frac{1}{2} [\cos(x - y) + 1]. \] Thus, the expression becomes: \[ 5 \sin x \sin y = \frac{5}{2} [\cos(x - y) + 1]. \]
Step 2: Maximize the expression To maximize the value of \( 5 \sin x \sin y \), we need to maximize \( \cos(x - y) \). The maximum value of \( \cos(x - y) \) is 1, which occurs when \( x = y \). Thus, when \( x = y \), the expression becomes: \[ 5 \sin x \sin y = \frac{5}{2} [1 + 1] = \frac{5}{2} \times 2 = 5. \] Therefore, the maximum value of \( 5 \sin x \sin y \) is 5.

The correct option is (C) : \(5\)

Answer 2

We are given that \(x\) and \(y\) are non-negative, \(x + y = \pi\), and we want to find the maximum value of \(5\sin x \sin y\).

Since \(x + y = \pi\), we can write \(y = \pi - x\). Substituting this into the expression \(5\sin x \sin y\), we get: \[5\sin x \sin(\pi - x)\]

Using the identity \(\sin(\pi - x) = \sin x\), we have: \[5\sin x \sin x = 5\sin^2 x\]

We want to maximize \(5\sin^2 x\). Since \(x\) and \(y\) are non-negative and \(x+y=\pi\), we have \(0 \le x \le \pi\). The maximum value of \(\sin x\) in this interval is 1, which occurs when \(x = \frac{\pi}{2}\). Therefore, the maximum value of \(\sin^2 x\) is also 1.

Thus, the maximum value of \(5\sin^2 x\) is \(5(1) = 5\).

Therefore, the maximum value of \(5\sin x \sin y\) is 5.