Profit Max? "A mill produces high & low quality steel. The profit on the former is double the latter...."

[Johnny V]

A mill produces high ( X ) & low ( Y ) quality steel . The profit on the former is double the latter . Given the production constraint Y = 98 - X / X , show that a total production of 20 maximizes profit .

I set up a profit function : P = 2x + y + L ( x + 98 - x / x ) . L is a Lagrange multiplier , P is profit . The co - efficients 1 & 2 are profit levels . I hoped to multiply out , take the first derivatives Px , Py , PL & solve the resulting 3 simultaneous equations . That's where I got as far as . Thank you .

 

[stapel]

A mill produces high ( X ) & low ( Y ) quality steel . The profit on the former is double the latter . Given the production constraint Y = 98 - X / X , show that a total production of 20 maximizes profit .

The expression "98 - X / X" means [imath]98 - \frac{X}{X}[/imath]. Did the text perhaps actually say "(98 - x)/x", which means [imath]\frac{98 - x}{x}[/imath]?

I set up a profit function : P = 2x + y + L ( x + 98 - x / x ) . L is a Lagrange multiplier , P is profit . The co - efficients 1 & 2 are profit levels . I hoped to multiply out , take the first derivatives Px , Py , PL & solve the resulting 3 simultaneous equations . That's where I got as far as .

Please provide a clear listing of the steps you mention having taken. Thank you.

Eliz.

 

[limiTS]

Let [imath]\displaystyle P=2x+\frac{98-x}{x}[/imath]
Differentiate [imath]P[/imath] with respect to [imath]x[/imath]
Set [imath]\displaystyle\frac{dP}{dx}=0[/imath]
Solve for [imath]x[/imath]
Then solve for [imath]\displaystyle y=\frac{98-x}{x}[/imath]
Does [imath]x+y=20[/imath]?
However, this may minimize the profit. Did you type in the question correctly?