An artist produces two items for sale. Each unit of item I costs $50 to produce, while each unit of item II costs $200. The revenue function is $R ( x , y ) = 40 x + 7 x y + 80 y ^ { 2 } + 10 y$ , where x is units of item I and y is units of item II. Suppose the artist has only $1000 to spend on production. Which of the following is the Lagrange function the artist should use to determine what combination of production amounts (x, y) will yield maximum profits subject to the constraint that his costs must equal $1000.
A) -10x + 7xy + 80 $y ^ { 2 }$ - 190y + λ(200x + 50y - 1000)
B) 40x + 7xy + 80 $y ^ { 2 }$ + 10y + λ(50x + 200y - 1000)
C) -10x + 7xy + 80 $y ^ { 2 }$ - 190y + λ(50x + 200y - 1000)
D) 40x + 7xy + 80 $y ^ { 2 }$ + 10y + λ(200x + 50y - 1000)
E) none of these

Question

Quizplus · Accepted Answer

The Lagrange function combines the objective function (in this case, the revenue function $R(x, y) = 40x + 7xy + 80y^2 + 10y$) with the constraint (the cost of producing x and y units, which must equal $1000) using a Lagrange multiplier (λ). The cost of producing x units of item I is $50x, and the cost of producing y units of item II is $200y. The constraint equation is $50x + 200y = 1000$. Therefore, the correct Lagrange function is $L(x, y, λ) = 40x + 7xy + 80y^2 + 10y + λ(50x + 200y - 1000)$.

An Artist Produces Two Items for Sale R(x,y)=40x+7xy+80y2+10yR ( x , y ) = 40 x + 7 x y + 80 y ^ { 2 } + 10 yR(x,y)=40x+7xy+80y2+10y

An Artist Produces Two Items for Sale $R ( x , y ) = 40 x + 7 x y + 80 y ^ { 2 } + 10 y$