Question

what is the maximum value of function y=3x²-y²+x³?

Answer 1

Answer:

hence the maximum value is 6 and minimum value is 2

Answer 2

The maximum value of function $f(x,y) = 3x^{2} - y^{2} + x^{3}$ is 4 at (-2, 0).

Step-by-step explanation:

Given: Function $f(x,y) = 3x^{2} - y^{2} + x^{3}$

To Find: The maximum value of the function

Solution:

• Finding the maximum value of function f(x,y) = 3x²-y²+x³

Given the function $f(x,y) = 3x^{2} - y^{2} + x^{3}$ such that

$\Rightarrow \dfrac{\partial f(x,y)}{\partial x} = \dfrac{\partial}{\partial x} (3x^{2} - y^{2} + x^{3}) = 6x + 3x^{2}$

$\Rightarrow \dfrac{\partial f(x,y)}{\partial y} = \dfrac{\partial}{\partial y} (3x^{2} - y^{2} + x^{3}) = -2y$

To find the maximum or minimum point, put $\Rightarrow \dfrac{\partial f(x,y)}{\partial x} = 0$

$\Rightarrow 6x + 3x^{2} = 0$ $\Rightarrow 3x (2 + x) = 0$

$\Rightarrow x = 0, -2$

And, $\Rightarrow \dfrac{\partial f(x,y)}{\partial y} = 0$

$\Rightarrow -2y = 0 \Rightarrow y = 0$

Therefore, the function $f(x,y)$ will be maximum at either of the two points, $(0,0)$ and $(-2, 0)$.

To determine which of the following points will have maximum value of the function $f(x,y)$, we have:  

$\Rightarrow r = \dfrac{\partial^2 f(x,y)}{\partial x^2} = 6 + 6x = 6(1+x)$

At the points (-2, 0) $\Rightarrow r |_{(-2,0)}= 6(1+(-2)) = -6 < 0$

At the points (0, 0) $\Rightarrow r |_{(0,0)}= 6(1+0) = 6 > 0$

Since $r < 0$ implies that the function is maximum at points (-2, 0), the maximum value is,

$\Rightarrow f(x,y)_{max} = 3(-2)^2 - (0)^2 + (-2)^3$

$\Rightarrow f(x,y)_{max} = 4$

Hence, the maximum value of function $f(x,y) = 3x^{2} - y^{2} + x^{3}$ is 4 at the point (-2, 0).