Question

The golf ball manufacturer has developed a profit model that depends on the number x of golf balls 

sold per month(measured in thousands),and the number of hours per month of advertising y, according 

to the function

𝑧 = 𝑓(𝑥, 𝑦) = 48𝑥 + 96𝑦 − 𝑥

2 − 2𝑥𝑦 − 9𝑦

2

,

Where z is measured in thousands of dollars .The budgetary constraint function relating the cost of the 

production of thousands golf balls and advertising units is given by 

20𝑥 + 4𝑦 = 216,

Find the values of x and y that maximize profit, and find the maximum profit

Answer 1

Given,

$f(x,y)=48x+96y-x^2-2xy-9y^2$

Constraints : $20x+4y=216=\phi(x,y)~~~~~~~~~-(1)$

By suing Lagrange's method-

$F(x,)=f(x,y)+\lambda \phi(x,y)$

    $=48x+96y-x^2-2xy-9y^2+\lambda(20x+4y-216)$

  

Now, $F'_x(x,y)=48-2x-2y+20\lambda=0~~~~~~~-(2)$

  $F'_y(X,y)=-96-2x-18y+4\lambda=0~~~~~~~$ -(3)

From equation 2 we have ,$\lambda=\dfrac{2x+2y-48}{20}$

    From eqn.(3) $\lambda=\dfrac{2x+18y-96}{4}$

Equating the above 2 values of $\lambda-$

$\dfrac{2x+2y-48}{205}=\dfrac{2x+18y-96}{4} \\[9pt] 2(x+y-24)=10(x+9y-48) \\[9pt] x+11y-54=0~~~~~-(4)$

Solving equation (1) and (4) and we get-

x=10,y=4

Hence, Maximum Profit

$Z=f(10,4)=48(10)+96(4)-10^2-2(10)(4)-9(4)^2 \\ =480+384-100-80-144\\ =540$