What is the Absolute Maximum of a Function on a Bounded Region?

[jegues]

Homework Statement

Find the maximum value of the function f(x, y) = xy(3 − x − 2y) on the triangle R bounded by the positive x- and y-axes and the line x+y = 1. Assume that f(x, y) has no critical points in the interior of R.

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

At the top of the page I have made a quick sketch of the triangle bounding the region R.

Since we are allowed to assume there are no critical points in R I proceeded to looking to the edges of the region for critical points.

I labeled them accordingly, C1, C2, C3.

On C1 we can see that x = 0 and it results in the function f(0,y) = 0.

To look for critical points on this edge we take the derivative and set it equal to 0 and solve.

Since 0 = 0 can I simply state that all the points on this edge are critical? This part confuses me.

The outcome is the same for C2 as well.

On C3 we can see that y=-x, resulsting in,

f(x,-x) = -x^{3} -3x^{2} = g(x)

I then took the derivative, factors and solved for the critical points which I found were at,

x=0,x=-2

When I plugged these values back into g(x) I obtained values of 0 and -4.

I don't really know what conclusion I can make without better classifying the edges C1 and C2, but as it stands right now I'd have to say that the absolute maximum occurs when Z=-4.

Can someone help clarify what I'm doing wrong or what I'm misunderstanding.

The answer listed is,

\frac{2 \sqrt{3}}{9}

Thanks again.

 

[Mark44]

Homework Statement

Find the maximum value of the function f(x, y) = xy(3 − x − 2y) on the triangle R bounded by the positive x- and y-axes and the line x+y = 1. Assume that f(x, y) has no critical points in the interior of R.

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

At the top of the page I have made a quick sketch of the triangle bounding the region R.

Since we are allowed to assume there are no critical points in R I proceeded to looking to the edges of the region for critical points.

I labeled them accordingly, C1, C2, C3.

On C1 we can see that x = 0 and it results in the function f(0,y) = 0.

To look for critical points on this edge we take the derivative and set it equal to 0 and solve.

Why bother taking the derivative? The function values are all zero along this edge.

Since 0 = 0 can I simply state that all the points on this edge are critical? This part confuses me.

The outcome is the same for C2 as well.

On C3 we can see that y=-x

No, on C3, y = 1 - x. This affects your subsequent work.

, resulsting in,

f(x,-x) = -x^{3} -3x^{2} = g(x)

I then took the derivative, factors and solved for the critical points which I found were at,

x=0,x=-2

When I plugged these values back into g(x) I obtained values of 0 and -4.

I don't really know what conclusion I can make without better classifying the edges C1 and C2, but as it stands right now I'd have to say that the absolute maximum occurs when Z=-4.

Can someone help clarify what I'm doing wrong or what I'm misunderstanding.

The answer listed is,

\frac{2 \sqrt{3}}{9}

Thanks again.

 

[jegues]

Thank you, I've got it now.