The minimum value of f(x, y)=x2-4 x+y2+6 y when x and y are subjected to the restrictions 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, is

Question

The minimum value of f(x, y)=x2-4 x+y2+6 y when x and y are subjected to the restrictions 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, is (A) - 1 (B) - 2 (C) - 3 (D) - 5

Tardigrade Admin · Accepted Answer

We have f ( x , y )= x 2+ y 2-4 x +6 y text Let ( x , y )=( cos θ, sin θ) text , then θ ∈[0, π / 2] text and f ( x , y )= f (θ)= cos 2 θ+ sin 2 θ-4 cos θ+6 sin θ f prime(θ)=6 cos θ+4 sin θ>0 ∀ θ ∈[0, π / 2] ∴ f prime(θ) text is strictly increasing in [0, π / 2] ∴ f prime(θ) min = f (0)=1-4+0=-3

Q. The minimum value of f(x,y)=x2−4x+y2+6y when x and y are subjected to the restrictions 0≤x≤1 and 0≤y≤1, is

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- 5

We have f(x,y)=x2+y2−4x+6y Let (x,y)=(cosθ,sinθ), then θ∈[0,π/2] and f(x,y)=f(θ)=cos2θ+sin2θ−4cosθ+6sinθ f′(θ)=6cosθ+4sinθ>0∀θ∈[0,π/2] ∴f′(θ) is strictly increasing in [0,π/2] ∴f′(θ)min​=f(0)=1−4+0=−3

Q. The minimum value of $f (x, y) = x^{2} - 4 x + y^{2} + 6 y$ when $x$ and $y$ are subjected to the restrictions $0 \leq x \leq 1$ and $0 \leq y \leq 1$ , is