# Math Help - Maximize

1. ## Maximize

Hi, I need some help with this problem:

Maximize Revenue given the following: R(x,y)=x(1000-0.1x2)+y(200-0.02y) subject to x>0, y>0, and 8x+3y<2400.

This is what I have so far but I don’t know what to do next.
Can some one help me please

▼R(x,y)=(1000-0.3x2)i+(200-0.04y)j
1000-0.3x2=0 --> x = ±√(1000/0.3)
200-0.04y=0 --> y = ±√(200/0.04)

2. Originally Posted by MarianaA
Hi, I need some help with this problem:

Maximize Revenue given the following: R(x,y)=x(1000-0.1x2)+y(200-0.02y) subject to x>0, y>0, and 8x+3y<2400.

This is what I have so far but I don’t know what to do next.
Can some one help me please

▼R(x,y)=(1000-0.3x2)i+(200-0.04y)j
1000-0.3x2=0 --> x = ±√(1000/0.3)
200-0.04y=0 --> y = ±√(200/0.04)
The second of these should be y=200/0.04=5000. Which gives the
critical points at (57.735,5000) and (-57.735,5000). Neither of these
are in the feasible region.

As the constraints are of strict inequality type there is no point in
the faesible region at which R(x,y) takes a maximal value.

If the constraints were:

x>=0, y>=0, and 8x+3y<=2400,

then the maximum would occur on the boundary of the feasible region, and
one could find it by examining R(x,y) on the boundary.

RonL

3. Originally Posted by CaptainBlack
The second of these should be y=200/0.04=5000. Which gives the
critical points at (57.735,5000) and (-57.735,5000). Neither of these
are in the feasible region.

As the constraints are of strict inequality type there is no point in
the faesible region at which R(x,y) takes a maximal value.

If the constraints were:

x>=0, y>=0, and 8x+3y<=2400,

then the maximum would occur on the boundary of the feasible region, and
one could find it by examining R(x,y) on the boundary.

RonL
but it is y = ± $\sqrt(200/0.04)$ not y=200/0.04
and how you get 57.735

4. Originally Posted by CaptainBlack
The second of these should be y=200/0.04=5000. Which gives the
critical points at (57.735,5000) and (-57.735,5000). Neither of these
are in the feasible region.

As the constraints are of strict inequality type there is no point in
the faesible region at which R(x,y) takes a maximal value.

If the constraints were:

x>=0, y>=0, and 8x+3y<=2400,

then the maximum would occur on the boundary of the feasible region, and
one could find it by examining R(x,y) on the boundary.

RonL
ok I undestand how you get 57.735, you get it by dividing $\sqrt(200/0.04)$ and 5000 by dividing 200/.04
But why?
Dont I need to substitute the values x = ± $\sqrt(1000/0.3)$, y = ± $\sqrt(200/0.04)$ the equation 8x+3y<2400
if it is this is what i get
8( $\sqrt(1000/0.3)$)+3( $\sqrt(200/0.04)$ )<2400 = 674.01

8(- $\sqrt(1000/0.3)$)+3( $\sqrt(200/0.04)$ )<2400 = -249.74

8( $\sqrt(1000/0.3)$)+3(- $\sqrt(200/0.04)$ )<2400 = 249.74

8(- $\sqrt(1000/0.3)$)+3(- $\sqrt(200/0.04)$ )<2400 = -674.01