Maximize z = 6x1 - 2x2+ 3x3 subject to 2x1 - x2 + 2x3 ≤ 2, x1+ 4x3 ≤ 4, x1, x2 , x3 ≥ 0.
Yes- a simple example is to think of two boxes with the same volume - 1cm3 the first a cube with sides 1cm and a surface area of 6 cm2 (6 equal sides) the second a box 2 X1 X1/2 the surface area is 2 sides of 2X1, 2 sides 1X1/2) and 2 sides 2X1/2 making in total 7 cm2
( x1 + x2) divided by 2 then (y1 +y 2) divided by 2
Rise divided by run. (Y2 - Y1) / (X2 - X1) - with (X1, Y1) and (X2, Y2) being two points on the graph.
5/8 x1/20 = 1/32
inverse matrix. x1+x2+x3=3,000 -x1+5x2=0 2x1-3x3=0.
Maximize z = 6x1 - 2x2+ 3x3 subject to 2x1 - x2 + 2x3 ≤ 2, x1+ 4x3 ≤ 4, x1, x2 , x3 ≥ 0.
3
this is the increasing function theorem, hope it helps "If F'(x) >= 0 , and all x's are and element of [a,b], Then F is increasing on [a,b]" use Mean Value Theorem (M.V.T) Let F'(x)>=0 on some interval Let x1< x2 (points from that interval) by M.V.T there is a point C which is an element of [x1,x2] such that F(x2)-F(x1) / X2- X1 = F'(C) this implies: F(x2)-F(x1) = F'(C) X [x2-x1] F'(C)>=0 [x2-x1]>0 therefore: F(x2)>=F(x1) Therefore: F is increasing on that interval.
Yes- a simple example is to think of two boxes with the same volume - 1cm3 the first a cube with sides 1cm and a surface area of 6 cm2 (6 equal sides) the second a box 2 X1 X1/2 the surface area is 2 sides of 2X1, 2 sides 1X1/2) and 2 sides 2X1/2 making in total 7 cm2
The degree of the polynomial 2x + 5 is 1. The highest power of x is x1, i.e. 2x1 + 5x0, hence the designation of first degree.
( x1 + x2) divided by 2 then (y1 +y 2) divided by 2
sqr.rtx/x= sqrt.x*sqr.rtx/sqr.rtx=x/x*sqrt.x=1/sqrt.x. x1/2 = x1/2 * x1/2 = x = 1 (x1/2) /x= 1/x1/2
formula for the midpoint of a line
m = y2 - y1 divided by x2 - x1
Rise divided by run. (Y2 - Y1) / (X2 - X1) - with (X1, Y1) and (X2, Y2) being two points on the graph.
x3/x1/2 = x5/2.