Proving Mean Value Inequality for sin(x) on 0≤x≤1 and 0≤y≤1

[absci2010]

Homework Statement

Show that 1/2(1-cos1)$$\leq$$$$\int$$$$\int$$sinx/(1+(xy)⁴)dxdy$$\leq$$1 on the area 0$$\leq$$x$$\leq$$1, 0$$\leq$$y$$\leq$$1.

Homework Equations

Mean Value Inequality: m*A(D)$$\leq$$$$\int$$$$\int$$f(x,y)dA$$\leq$$M*A(D), where m is the minimum and M is the maximum on the interval.

The Attempt at a Solution

A(D)=1
sinx$$\leq$$1, sinx/(1+(xy)⁴)$$\leq$$1, so M=1
I tried to find the minimum on the interval, but I can't find any critical points that are in 0$$\leq$$x$$\leq$$1 and 0$$\leq$$y$$\leq$$1.
Please help?

 

[Tedjn]

I have not worked this question out, but I suspect you need something more than mA(D) as a minimum. When you plug in x = 0 and y = anything in D, you find a minimum value of 0.

 

[fzero]

If there are no critical points, the minimum should occur somewhere on the boundary.

 

[absci2010]

Yes, but that's the part I'm confused about. Where does the value (1/2)(1-cos1) come from?

 

[fzero]

Yes, but that's the part I'm confused about. Where does the value (1/2)(1-cos1) come from?

If you look hard enough at that quantity, you might be able to guess at a $$g(x,y)$$, such that

$$g(x,y)\leq f(x,y), ~ (x,y)\in D,$$

for which the integration is elementary.