Question

Найти наибольшее и наименьшее значения функции z = 4x2 + y2 −2 y при x ≤1 , 0 ≤ y − x ≤1 , 0 ≤ x + y ≤1.

Answer 1

Ответ:

minz(x;y)=-1; maxz(x;y)=1

Пошаговое объяснение:

z(x;y)=4x²+y²−2y  при x ≤1 , 0 ≤ y − x ≤1 , 0 ≤ x + y ≤1

1) 0 ≤ y − x ≤1

  0 ≤ x + y ≤1

0+0 ≤ (y − x)+(x + y) ≤1+1 ⇒ 0 ≤ y ≤1

2) 0 ≤ y − x ≤1⇒-1≤ x-y≤0

-1≤ x-y≤0

0 ≤ x + y ≤1

-1+0≤ (x-y)+(x+y)≤0+1⇒-0,5≤ x≤0,5

3) z(x;y)=4x²+y²−2y=4x²+(y-1)²-1

-0,5≤ x≤0,5 ⇒ 0≤4x²≤1

0 ≤ y ≤1 ⇒ 0≤(y-1)²≤1

0+0-1≤4x²+(y-1)²-1≤1+1-1

-1≤4x²+(y-1)²-1≤1

minz(x;y)=z(0;1)=-1

maxz(x;y)=z(0,5;0)=1

Answer 2

To find the largest and smallest values of the function $z = 4x^2 + y^2 - 2y$ subject to the given constraints, we need to examine the critical points within the domain defined by the inequalities $x \leq 1$, $0 \leq y - x \leq 1$, and $0 \leq x + y \leq 1$.

Let's start by identifying the critical points. A critical point occurs where the gradient of the function is equal to zero or is undefined. In this case, the gradient is defined as:

$\nabla z = \left(\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}\right)$

So, let's find the partial derivatives of $z$ with respect to $x$ and $y$:

$\frac{\partial z}{\partial x} = 8x$ $\frac{\partial z}{\partial y} = 2y - 2$

Next, we set these partial derivatives equal to zero and solve for $x$ and $y$:

For $\frac{\partial z}{\partial x} = 8x = 0$, we get $x = 0$.

For $\frac{\partial z}{\partial y} = 2y - 2 = 0$, we get $y = 1$.

Thus, the critical point is $(x, y) = (0, 1)$.

Now, let's evaluate the function $z$ at the critical point and at the boundaries of the domain to find the maximum and minimum values:

1. Critical Point: $z(0, 1) = 4 \cdot 0^2 + 1^2 - 2 \cdot 1 = -1$.

2. Boundaries: a. $x = 0$: $z(0, y) = 4 \cdot 0^2 + y^2 - 2y = y^2 - 2y$. b. $x = 1$: $z(1, y) = 4 \cdot 1^2 + y^2 - 2y = 4 + y^2 - 2y$. c. $y - x = 0$: $z(x, x) = 4x^2 + x^2 - 2x = 5x^2 - 2x$. d. $y - x = 1$: $z(x, x + 1) = 4x^2 + (x + 1)^2 - 2(x + 1) = 5x^2 + 1$. e. $x + y = 1$: $z(x, 1 - x) = 4x^2 + (1 - x)^2 - 2(1 - x) = 5x^2 - 2x + 1$.

Now, we need to compare the values of $z$ at these points to find the maximum and minimum values:

• The maximum value will be the largest of all the values found.
• The minimum value will be the smallest of all the values found.

Let's analyze each case:

1. Critical Point: $z = -1$.

2. Boundaries: a. $z = y^2 - 2y$. b. $z = 4 + y^2 - 2y$. c. $z = 5x^2 - 2x$. d. $z = 5x^2 + 1$. e. $z = 5x^2 - 2x + 1$.

To find the maximum and minimum values, we should consider the values of $z$