Find the maximum value of the function $f ( x , y ) = x ^ { 2 } + 4 y$ subject to $x ^ { 2 } + y ^ { 2 } = 1$
A) 4
B) -4
C) 1
D) -1

Question

Quizplus · Accepted Answer

The constraint $x^2 + y^2 = 1$ represents a circle with radius 1 centered at the origin. To find the maximum value of $f(x, y) = x^2 + 4y$, we can use the method of Lagrange multipliers or directly substitute $y = \sqrt{1 - x^2}$ (considering only the positive square root for the maximum y, as the function increases with y). The derivative method would show that the function $f(x, y)$ reaches its maximum when $y$ is maximized, which happens at the top of the circle where $y = 1$. Substituting $y = 1$ into $f(x, y)$ gives $f(x, 1) = x^2 + 4$. Since $x^2$ is always positive or zero, the maximum value of $f(x, y)$ on the given circle is achieved when $x^2$ is minimized, which is 0. Therefore, the maximum value of $f(x, y)$ is $0 + 4 = 4$.

Find the Maximum Value of the Function f(x,y)=x2+4yf ( x , y ) = x ^ { 2 } + 4 yf(x,y)=x2+4y

Find the Maximum Value of the Function $f ( x , y ) = x ^ { 2 } + 4 y$