Application of DerivativeshardFree

Maximum of x - y Subject to x^2 + y^2 = 27 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
If xx and yy are real numbers such that x2+y2=27x^{2}+y^{2} = 27, then the maximum possible value of x−yx-y is
A72\sqrt{72}
B18\sqrt{18}
C54\sqrt{54}correct
D23\sqrt{\dfrac23}
Solution
Step 1: Use the constraint to parametrise. Any point on the circle x2+y2=27x^{2}+y^{2} = 27 can be written as
x=27cos⁡θ,y=27sin⁡θ,θ∈[−π, π].x = \sqrt{27}\cos\theta, \qquad y = \sqrt{27}\sin\theta, \qquad \theta \in [-\pi,\ \pi] .
 Step 2: Write the quantity to be maximised.
A=x−y=27(cos⁡θ−sin⁡θ).A = x - y = \sqrt{27}\left(\cos\theta - \sin\theta\right).
 Step 3: Bound the bracket. For pcos⁡θ+qsin⁡θp\cos\theta + q\sin\theta the range is [−p2+q2, p2+q2]\left[-\sqrt{p^{2}+q^{2}},\ \sqrt{p^{2}+q^{2}}\right]; here p=1p = 1, q=−1q = -1, so
−2 ≤ cos⁡θ−sin⁡θ ≤ 2.-\sqrt2 \ \le\ \cos\theta - \sin\theta \ \le\ \sqrt2 .
 Step 4: Take the largest value.
Amax⁡=27⋅2=54.A_{\max} = \sqrt{27}\cdot\sqrt2 = \sqrt{54} .
 Step 5: Check it is attained. Equality needs cos⁡θ−sin⁡θ=2\cos\theta - \sin\theta = \sqrt2, i.e. θ=−π4\theta = -\dfrac{\pi}{4}, giving
x=27⋅12=332,y=−332,x = \sqrt{27}\cdot\frac{1}{\sqrt2} = \frac{3\sqrt3}{\sqrt2}, \qquad y = -\frac{3\sqrt3}{\sqrt2} ,
 and then x2+y2=272+272=27x^{2}+y^{2} = \dfrac{27}{2}+\dfrac{27}{2} = 27 with x−y=632=54x - y = \dfrac{6\sqrt3}{\sqrt2} = \sqrt{54}. Answer: (3).
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