Profitable Golf Balls

Figure 1.
Pro-T company 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−x2−2xy−9y2
where z is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is 50,000, and the maximum number of hours of advertising that can be purchased is 25. Find the values of x and y that maximize profit, and find the maximum profit.
Solution
Using the problem-solving strategy, step 1 involves finding the critical points of f on its domain. Therefore, we first calculate fx(x,y) and fy(x,y), then set them each equal to zero:
fx(x,y)=48−2x−2yfy(x,y)=96−2x−18y
Setting them equal to zero yields the system of equations
48−2x−2y=096−2x−18y=0
The solution to this system is x=21 and y=3. Therefore (21,3) is a critical point of f. Calculating f(21,3) gives f(21,3)=48(21)+96(3)−212−2(21)(3)−9(3)2=648.
The domain of this function is 0≤x≤50 and 0≤y≤25 as shown in the following graph.

Figure 2. Graph of the domain of the functionz=f(x,y)=48x+96y−x2−2xy−9y2.
L1 is the line segment connecting (0,0) and (50,0), and it can be parameterized by the equations x(t)=t, y(t)=0 for 0≤t≤50. We then define g(t)=f(x(t),y(t)):
g(t)=f(x(t),y(t))=f(t,0)=48t+96(0)−y2−2(t)(0)−9(0)2=48t−t2
Setting g′(t)=0 yields the critical point t=24, which corresponds to the point (24,0) in the domain of f. Calculating f(24,0) gives 576.
L2 is the line segment connecting (50,0) and (50,25), and it can be parameterized by the equations x(t)=50, y(t)=t for 0≤t≤25. Once again we define g(t)=f(x(t),y(t)):
g(t)=f(x(t),y(t))=f(50,t)=48(50)+96t−502−2(50)t−9t2=−9t2−4t−100
This function has a critical point t=−29, which corresponds to the point (50,−29). This point is not in the domain of f.
L3 is the line segment connecting (0,25) and (50,25), and it can be parameterized by the equations x(t)=t, y(t)=25 for 0≤t≤50. We define g(t)=f(x(t),y(t)):
g(t)=f(x(t),y(t))=f(t,25)=48(t)+96(25)−t2−2t(25)−9(25)2=−t2−2t−3225
This function has a critical point t=−1, which corresponds to the point (−1,25), which is not in the domain.
L4 is the line segment connecting (0) and (0,25), and it can be parameterized by the equations x(t)=0, y(t)=t for 0≤t≤25. We define g(t)=f(x(t),y(t)):
g(t)=f(x(t),y(t))=f(0,t)=48(0)+96t−(0)2−2(0)t−9t2=96t−t2
This function has a critical point t=163, which corresponds to the point (0,163), which is on the boundary of the domain. Calculating f(0,163) gives 256.
We also need to find the values of f(x,y) at the corners of its domain. These corners are located at (0,0),(50,0),(50,25), and (0,25):
f(0,0)=48(0)+96(0)−(0)2−2(0)(0)−9(0)2=0f(50,0)=48(50)+96(0)−(50)2−2(50)(0)−9(0)2=−100f(50,25)=48(50)+96(25)−(500)2−2(50)(25)−9(25)2=−5825f(0,25)=48(0)+96(25)−(0)2−2(0)(25)−9(25)2=−3225
The maximum critical value is 648 which occurs at (21,3). Therefore, a maximum profit of $648,000 is realized when 21,000 golf balls are sold and 33 hours of advertising are purchased per month as shown in the following figure.

Figure 3. The profit function f(x,y) has a maximum at (21,3,648).
Candela Citations
- Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: https://openstax.org/books/calculus-volume-3/pages/1-introduction. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction