Given that ๐‘“(๐‘ฅ) = 3 cos ๐‘ฅ + 2 sin ๐‘ฅ, where 0 โ‰ค

๐‘ฅ โ‰ค 90, for what value of ๐‘ฅ ๐‘–๐‘  ๐‘“(๐‘ฅ) maximum?
A. 33.7
B. 56.3
C. 213.7
D. 236.3

Bot Bot answered

We can use calculus to find the maximum value of ๐‘“(๐‘ฅ). Taking the derivative of ๐‘“(๐‘ฅ), we get:

๐‘“'(๐‘ฅ) = -3 sin ๐‘ฅ + 2 cos ๐‘ฅ

Setting ๐‘“'(๐‘ฅ) = 0 to find critical points, we get:

-3 sin ๐‘ฅ + 2 cos ๐‘ฅ = 0

Dividing both sides by cos ๐‘ฅ, we get:

-3 tan ๐‘ฅ + 2 = 0

Solving for tan ๐‘ฅ, we get:

tan ๐‘ฅ = 2/3

Using a calculator, we find that ๐‘ฅ โ‰ˆ 33.7ยฐ or ๐‘ฅ โ‰ˆ 213.7ยฐ. We can check these values to see which one gives us the maximum value of ๐‘“(๐‘ฅ).

At ๐‘ฅ = 33.7ยฐ, we have:

๐‘“(33.7ยฐ) = 3 cos 33.7ยฐ + 2 sin 33.7ยฐ โ‰ˆ 3.29

At ๐‘ฅ = 213.7ยฐ, we have:

๐‘“(213.7ยฐ) = 3 cos 213.7ยฐ + 2 sin 213.7ยฐ โ‰ˆ -0.26

Therefore, the value of ๐‘ฅ for which ๐‘“(๐‘ฅ) is maximum is ๐‘ฅ โ‰ˆ 33.7ยฐ, which is closest to option A.

To find the value of ๐‘ฅ for which ๐‘“(๐‘ฅ) is maximum, we need to find the maximum value of the function ๐‘“(๐‘ฅ) = 3 cos ๐‘ฅ + 2 sin ๐‘ฅ.

๐‘“(๐‘ฅ) = 3 cos ๐‘ฅ + 2 sin ๐‘ฅ can be written as ๐‘“(๐‘ฅ) = ๐‘Ÿ sin(๐‘ฅ + ๐œƒ), where ๐‘Ÿ is the magnitude of the vector and ๐œƒ is the angle.

We can rewrite ๐‘“(๐‘ฅ) = 3 cos ๐‘ฅ + 2 sin ๐‘ฅ as ๐‘“(๐‘ฅ) = 3 (cos ๐‘ฅ/โˆš(๐‘ก^2 + 3^2)) + 2 (sin ๐‘ฅ/โˆš(๐‘ก^2 + 2^2)), where ๐‘ก = tan(๐œƒ/2).

The maximum value of the function ๐‘“(๐‘ฅ) occurs when ๐‘ก is maximum, which means when ๐œƒ is maximum.

The maximum value of ๐‘ก is given by ๐‘ก = tan(๐œƒ_max/2) = โˆš(9/4) = 3/2.

To find the corresponding angle ๐œƒ_max, we use the inverse tangent function: ๐œƒ_max = 2 * atan(3/2).

๐œƒ_max โ‰ˆ 56.31 degrees.

Therefore, for ๐‘ฅ = ๐œƒ_max, ๐‘“(๐‘ฅ) is maximum.

The answer is B. 56.3.