Question

Find the maximum and minimum values of f(x,y) = 2+2x+4y-x^2-y^2 on the triangular region in the first quadrant bounded by the lines x=0, y0, y=9x.

Answer 1

I think your question seems like :-

Find the maximum and minimum values of f(x,y) = 2+2x+4y-x²-y² on the triangular region in the first quadrant bounded by the lines x = 0, y = 0, y = 9-x. The reason is below :

If you see carefully, the given lines were x = 0, y = 0, y = 9x , also these lines forms a Triangle in first quadrant (Given) , but these all lines doesn't form a triangle in 1st quadrant , we can't change the other two lines (As those two lines will work as permanent sides and support for the triangle) but third one can be , also neither 9 + x nor (9/x) can't form triangle in 1st quadrant with the given other two lines .Only with y = 9 - x the given situation can be achieved

But before starting the question let's recall some of the concepts :-

For a given function let it be f(x,y) ,

• f(x,y) can have maximum or minimum if AC - B² > 0 , as per A < 0 and A > 0 , if A < 0 then maxima at critical points and if A > 0 , then minima at critical points
• f(x,y) have saddle points if AC - B² < 0
• From AC - B² = 0 , nothing can be concluded

Where ,

• ${\boxed{\bf{A=\dfrac{\partial^{2}f}{\partial x^{2}}}}}$

• ${\boxed{\bf{B=\dfrac{\partial^{2}f}{\partial x\partial y}}}}$

• ${\boxed{\bf{C=\dfrac{\partial^{2}f}{\partial y^{2}}}}}$

Now , consider the function ;

${:\implies \quad \sf f(x,y)=2+2x+4y-x^{2}-y^{2}}$

Partial Differentiating both sides w.r.t.x

${:\implies \quad \sf \dfrac{\partial f}{\partial x}=2-2x}$

Now , partial differentiating both sides w.r.t.y

${:\implies \quad \sf \dfrac{\partial^{2}f}{\partial x\partial y}=0}$

Now , consider ;

${:\implies \quad \sf \dfrac{\partial f}{\partial x}=2-2x}$

Partial differentiating both sides w.r.t.x

${:\implies \quad \sf \dfrac{\partial^{2}f}{\partial x^{2}}=-2}$

Now , consider ;

${:\implies \quad \sf f(x,y)=2+2x+4y-x^{2}-y^{2}}$

Partial differentiating both sides w.r.t.y

${:\implies \quad \sf \dfrac{\partial f}{\partial y}=4-2y}$

Now , partial differentiating both sides w.r.t.y

${:\implies \quad \sf \dfrac{\partial^{2}f}{\partial y^{2}}=-2}$

Now to find the critical points , first order partial derivative of f(x,y) both w.r.t.x and w.r.t.y must be 0. So , if you equate them to 0 , you will get x = 1 and y = 2. So critical point is only one i.e (1,2)

Now , from the above conditions , let calculate AC - B² first , from which we will get maxima or minima

${:\implies \quad \sf (-2)(-2)-(0)^{2}}$

${:\implies \quad \sf 4\> 0}$

So , AC - B² > 0 and also A < 0 , so maxima at (1,2) , to find maxima we need to put x = 1 , y = 2 in f(x,y) i.e f(1,2) = 2 + 2(1) + 4(2) - 1² - 2² = 7 . Now , for minima , draw the triangle see attachment 1 . Now , in the attachment critical points are (0,9) , (0,0) and (9,0) . Now , we need to find f(x,y) at these all points too . If you calculate then you will get f(0,9) = -43 , f(0,0) = 2 and f(9,0) = -61 , out of which -61 is minimum

So , the required maxima and minima are 7 and -61

Note :- Also , refer to the attachments no. 2,3,4 as well for why I choosed the third line to be 9 - x rather than others.