The function given is:

$f(x, y) = 2x^3 - 5y$

with the constraints $0 \leq x \leq 1$ and $0 \leq y \leq 1$. To find the global extreme values of this function, we will follow these steps:

Step 1: Find Critical Points

We begin by finding the partial derivatives of the function with respect to $x$ and $y$, and then set them equal to zero to find critical points.

$\frac{\partial f}{\partial x} = 6x^2$ $\frac{\partial f}{\partial y} = -5$

Setting these derivatives equal to zero:

$6x^2 = 0 \quad \Rightarrow \quad x = 0$ -5 = 0 \quad \text{(This is never true, so no critical points with respect to y.)}

Thus, the only critical point occurs at $x = 0$.

Step 2: Evaluate the Function at the Boundary Points

Since the region is constrained by $0 \leq x \leq 1$ and $0 \leq y \leq 1$, we need to evaluate the function at the boundary points of the region.

Evaluate at the corners of the region:

1. $f(0, 0) = 2(0)^3 - 5(0) = 0$
2. $f(0, 1) = 2(0)^3 - 5(1) = -5$
3. $f(1, 0) = 2(1)^3 - 5(0) = 2$
4. $f(1, 1) = 2(1)^3 - 5(1) = 2 - 5 = -3$

Step 3: Identify the Global Maximum and Minimum

From the evaluations at the critical and boundary points:

• $f(0, 0) = 0$
• $f(0, 1) = -5$
• $f(1, 0) = 2$
• $f(1, 1) = -3$

Thus, the global maximum value is $\boxed{2}$ at $(1, 0)$, and the global minimum value is $\boxed{-5}$ at $(0, 1)$.

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5 Related Questions:

1. How do partial derivatives help in finding critical points?
2. What are global extreme values and how do they differ from local extreme values?
3. How would the solution change if the function were different, say $f(x, y) = x^2 + y^2$?
4. How does one verify if a critical point is a maximum or minimum?
5. How do boundary conditions affect the global extrema of a function?

Tip: Always check both critical points and boundary points when determining global extrema on a closed region.