# Math Help - maximum percentage error

1. ## maximum percentage error

I have this function again:

$
f(x,y) = x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6
$

The value of $f$ when $x = -1$ and $y = 2$ is $25$.

The maximum percentage error in the value of $x$ is $2$% and the maximum percentage error in the value of $y$ is $3$%.

I have to find the least and greatest possible values that $f(-1,2)$ can take, and the maximum possible percentage error in taking the value $f(-1,2) = 25$.

I plugged $-1$ and $2$ into the partial derivatives:

$
f$
x $(-1,2) = 4(-1)[(-1)^2 + (2)^2 + 1] = -24
$

$
f$
y $(-1,2) = 4(2)[(2)^2 + (-1)^2 - 1] = 32
$

And then found:

$
\delta f = -24(0.02)(-1) + 32(0.03)(2) = 2.4
$

So the least and greatest possible values for $f(-1,2)$ are $22.6$ and $27.4$, and the maximum possible percentage error is $9.6$% ?

Is there anything correct here or am I completely wrong?

2. Originally Posted by billym
I have this function again:

$
f(x,y) = x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6
$

The value of $f$ when $x = -1$ and $y = 2$ is $25$.

The maximum percentage error in the value of $x$ is $2$% and the maximum percentage error in the value of $y$ is $3$%.

I have to find the least and greatest possible values that $f(-1,2)$ can take, and the maximum possible percentage error in taking the value $f(-1,2) = 25$.

I plugged $-1$ and $2$ into the partial derivatives:

$
f$
x $(-1,2) = 4(-1)[(-1)^2 + (2)^2 + 1] = -24
$

$
f$
y $(-1,2) = 4(2)[(2)^2 + (-1)^2 - 1] = 32
$

And then found:

$
\delta f = -24(0.02)(-1) + 32(0.03)(2) = 2.4
$

So the least and greatest possible values for $f(-1,2)$ are $22.6$ and $27.4$, and the maximum possible percentage error is $9.6$% ?

Is there anything correct here or am I completely wrong?
What is wrong with your method is that you are approximating the change in the function by using the derivative at (-1, 2), but the function values you want to calculate are not at (-1, 2).

I would find out if the derivative is equal to zero on the area of interest and the simply use the Extreme Value Theorem with endpoints x = -1.03, x = -.97, y = 1.96, y = 2.04. The minimum and maximum values of the function on that area must either occur where the derivative is zero or on the border.

3. The question says I have to use first-order partial derivatives to determine the values, but I can't find a single example of this kind of question in my notes.

I have no idea what I am doing.

Do 24.52 and 25.48 make any more sense?

4. Originally Posted by billym
The question says I have to use first-order partial derivatives to determine the values, but I can't find a single example of this kind of question in my notes.

I have no idea what I am doing.

Do 24.52 and 25.48 make any more sense?
I can't think of what you would be using the partial derivatives for except checking to see if $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$ at some point in the area of interest.