`S={3x^2 le 4y le 6x+2y}` find the area

Draw a rough sketch of the region {(x,y) : y^(2)le 6 a x and x^(2)+y^(2)le 16 a^(2)} . Also, find the area of the region sketched using method of integration.

x + y le 6, x + y le 4

The area of region between y=4-x^(2), 0 le x le 2 and the x-axis is

The maximum and minimum values of the objective function Z = 3x - 4 y subject to the constraints x- 2y le 0, -3x +y le 4 x-y le 6 , x,y ge 0 are respectively

If 1 le x le 3 and 2 le y le 4 , what is the maximum value of ((x)/(y)) ?

The area of the region : R = {(x, y) : 5x^2 le y le 2x^2 + 9} is :