Solution to the Problem:

The prices are: $1.20, $1.25, $1.50, and $3.16

Let's begin by representing the problem with a pair of equations.
x y z w = 7.11
x + y + z + w = 7.11

If x,y,z,w represent the amounts of the four items in cents then we have

x y z w = 711000000
x + y + z + w = 711

What can we say so far?

The factors : 711 = (1)(3)(3)(79)

If there is to be an exact solution then the product xyzw must be divisible by 3, 9, 79.
The exact sum of the 4 items is 7.11 so the cost of each item is less than $7.11.

SOLUTION #1 (exact solution)
Suppose the customer purchased four items worth $3.16, $1.50, $1.20, and $1.25,
then the total cost and the product would be exactly $7.11 .

SOLUTION #2 (rounds correctly)
Suppose the customer selected four items which cost $.79, $1.75, $2.00, $2.57.
The total cost is $7.11. However, the product is 7.10605.
The clerk rounds correctly and quotes the amount $7.11.

SOLUTION #3 (rounds correctly)
Suppose the customer selected four items which cost $2.00, $2.07, $2.29, $.75.
The total cost is $7.11. However, the product is 7.11045.
The clerk rounds correctly and quotes the amount $7.11.

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Mikael Wetterholm sent in the following:
I wrote a simple BASIC-program and in ten seconds I had the answer on my screen: $1.2 $1.25 $1.5 $3.16.

CLS
sum = 711 'sum in cent
product = 711000000 'product in cent
a = 100' I suggested that the cost of an item was greater than 100 cent

PRINT "wait..."

DO WHILE a < sum
b = 100
a = a + 1
     DO WHILE b < sum
     c = 100
     b = b + 1
          DO
          c = c + 1
          d = sum - (a + b + c)
         IF a * b * c * d = product THEN GOTO final
         LOOP WHILE d > 100
     LOOP
LOOP
END

final:
a = a / 100 'convert to dollar
b = b / 100
c = c / 100
d = d / 100
SOUND 1000, 10
PRINT "$"; a; "$"; b; "$"; c; "$"; d;
END

Dave Smith also wrote a computer program to solve the problem.