Maximizing Profit for a Metal Manufacturing Company with Limited Budget

[jmoney1014]

[I've been on this problem for hours :( Someone please help.

A metal manufacturing company estimates that if x million dollars is spent on research and development (R&D) and y million dollars is spent on marketing, the total revenue from a material will be R=45x^0.04y^0.12 million dollars.

If the metal manufacturing company has exactly $50 million to devote to this new material, how should they allocate the money between R&D and marketing in order to maximize their profit?

What will this maximum profit be?


I'm just not sure how to approach this, it seems like such an easy problem. I know that x and y are non negative so I try to solve the constraint equation for one of the variables but I keep falling into a hole, any help would be appreciated

 

[Mark44]

[I've been on this problem for hours :( Someone please help.

A metal manufacturing company estimates that if x million dollars is spent on research and development (R&D) and y million dollars is spent on marketing, the total revenue from a material will be R=45x^0.04y^0.12 million dollars.

If the metal manufacturing company has exactly $50 million to devote to this new material, how should they allocate the money between R&D and marketing in order to maximize their profit?

What will this maximum profit be?


I'm just not sure how to approach this, it seems like such an easy problem. I know that x and y are non negative so I try to solve the constraint equation for one of the variables but I keep falling into a hole, any help would be appreciated

You have R(x, y) = 45x^.04y^.12 (million dollars)

You can write R as a function of one variable by using the fact that $50 million is to be spent on R&D and marketing.

 

[jack action]

x+y=50

To find x where R(x) is maximum: dR(x) / dx = 0 ===> Isolate x

[or to find y where R(y) is maximum: dR(y) / dy = 0 ===> Isolate y]

If have difficulties seeing the problem, plot the function R(x). You will see that at the value of x where R(x) is maximum, the derivative dR(x) / dx = 0, i.e. the slope is an horizontal line at that point.