diff --git a/docs/src/tutorials/linear/diet.jl b/docs/src/tutorials/linear/diet.jl
index 701d3e7248d..d8f713413bd 100644
--- a/docs/src/tutorials/linear/diet.jl
+++ b/docs/src/tutorials/linear/diet.jl
@@ -79,7 +79,7 @@ limits = DataFrames.DataFrame(
         "fat" 0 65
         "sodium" 0 1779
     ],
-    ["name", "min", "max"],
+    ["nutrient", "min", "max"],
 )
 
 # ## JuMP formulation
@@ -90,32 +90,30 @@ limits = DataFrames.DataFrame(
 # HiGHS as our optimizer:
 
 model = Model(HiGHS.Optimizer)
+set_silent(model)
 
-# Next, we create a set of decision variables `x`, indexed over the foods in the
-# `data` DataFrame. Each `x` has a lower bound of `0`.
+# Next, we create a set of decision variables `x`, with one element for each row
+# in the DataFrame, and each `x` has a lower bound of `0`:
 
-@variable(model, x[foods.name] >= 0);
+@variable(model, x[foods.name] >= 0)
 
-# Our objective is to minimize the total cost of purchasing food. We can write
-# that as a sum over the rows in `data`.
+# To simplify things later on, we store the vector as a new column `x` in the
+# DataFrame `foods`:
 
-@objective(
-    model,
-    Min,
-    sum(food["cost"] * x[food["name"]] for food in eachrow(foods)),
-);
+foods.x = Array(x)
+
+# Our objective is to minimize the total cost of purchasing food:
+
+@objective(model, Min, sum(foods.cost .* foods.x));
 
 # For the next component, we need to add a constraint that our total intake of
-# each component is within the limits contained in the `limits` DataFrame.
-# To make this more readable, we introduce a JuMP `@expression`
-
-for limit in eachrow(limits)
-    intake = @expression(
-        model,
-        sum(food[limit["name"]] * x[food["name"]] for food in eachrow(foods)),
-    )
-    @constraint(model, limit.min <= intake <= limit.max)
-end
+# each component is within the limits contained in the `limits` DataFrame:
+
+@constraint(
+    model,
+    [row in eachrow(limits)],
+    row.min <= sum(foods[!, row.nutrient] .* foods.x) <= row.max,
+);
 
 # What does our model look like?
 
@@ -132,8 +130,8 @@ solution_summary(model)
 
 # Success! We found an optimal solution. Let's see what the optimal solution is:
 
-for food in foods.name
-    println(food, " = ", value(x[food]))
+for row in eachrow(foods)
+    println(row.name, " = ", value(row.x))
 end
 
 # That's a lot of milk and ice cream! And sadly, we only get `0.6` of a
@@ -155,7 +153,9 @@ filter!(row -> row.quantity > 0.0, solution)
 # constraints. Let's see what happens if we add a constraint that we can buy at
 # most 6 units of milk or ice cream combined.
 
-@constraint(model, x["milk"] + x["ice cream"] <= 6)
+dairy_foods = ["milk", "ice cream"]
+is_dairy = map(name -> name in dairy_foods, foods.name)
+@constraint(model, sum(foods[is_dairy, :x]) <= 6)
 optimize!(model)
 Test.@test termination_status(model) == INFEASIBLE  #hide
 Test.@test primal_status(model) == NO_SOLUTION      #hide