15. [-/2.5 Points] DETAILS MY NOTES WANEFMAC8 6.2.029. Gerber Multigrain Cereal contains, in each serving, 60 calories and 11 grams of carbohydrates. Gerber 2nd Foods Mango contains, in each serving, 80 calories and 19 grams of carbohydrates. If the cereal costs $$ 30 ¢ $$ per serving and the dessert costs $$ 52 ¢ $$ per serving, and you want to provide your child with at least 140 calories and at least 30 grams of carbohydrates, how can you do so at the least cost? (Fractions of servings are permitted. If an answer does not exist, enter DNE.) Gerber Multigrain Cereal Gerber 2nd Foods Mango

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UpStudy AI · Accepted Answer

To solve this problem, we need to set up a system of inequalities based on the given conditions and then minimize the cost function.

### Known Conditions:
1. **Gerber Multigrain Cereal:**
   - Calories per serving: $$ 60 $$
   - Carbohydrates per serving: $$ 11 $$
   - Cost per serving: $$ 0.30 $$

2. **Gerber 2nd Foods Mango:**
   - Calories per serving: $$ 80 $$
   - Carbohydrates per serving: $$ 19 $$
   - Cost per serving: $$ 0.52 $$

### Variables:
Let:
- $$ x $$ = number of servings of Gerber Multigrain Cereal
- $$ y $$ = number of servings of Gerber 2nd Foods Mango

### Inequalities:
We need to satisfy the following conditions:
1. Total calories: 
   $$
   60x + 80y \geq 140
   $$
2. Total carbohydrates: 
   $$
   11x + 19y \geq 30
   $$

### Cost Function:
We want to minimize the cost:
$$
C = 0.30x + 0.52y
$$

### Setting Up the System:
We will solve the inequalities and then find the minimum cost. Let's first express the inequalities in standard form and then solve them.

1. $$ 60x + 80y \geq 140 $$
2. $$ 11x + 19y \geq 30 $$

Now, I will solve this system of inequalities to find the feasible region.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}60x+80y=140\11x+19y=30\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{7-4y}{3}\11x+19y=30\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$11\times \frac{7-4y}{3}+19y=30$$
- step3: Simplify:
  $$\frac{11\left(7-4y\right)}{3}+19y=30$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{11\left(7-4y\right)}{3}+19y\right)\times 3=30\times 3$$
- step5: Simplify the equation:
  $$77+13y=90$$
- step6: Move the constant to the right side:
  $$13y=90-77$$
- step7: Subtract the numbers:
  $$13y=13$$
- step8: Divide both sides:
  $$\frac{13y}{13}=\frac{13}{13}$$
- step9: Divide the numbers:
  $$y=1$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{7-4\times 1}{3}$$
- step11: Substitute back:
  $$x=\frac{7-4}{3}$$
- step12: Calculate:
  $$x=1$$
- step13: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=1\end{array}\right.$$
- step14: Check the solution:
  $$\left\{ \begin{array}{l}x=1\y=1\end{array}\right.$$
- step15: Rewrite:
  $$\left(x,y\right) = \left(1,1\right)$$
Solve the system of equations $$ 60x + 80y = 140; 11x + 19y = 30 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}60x+80y=140\11x+19y=30\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{7-4y}{3}\11x+19y=30\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$11\times \frac{7-4y}{3}+19y=30$$
- step3: Simplify:
  $$\frac{11\left(7-4y\right)}{3}+19y=30$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{11\left(7-4y\right)}{3}+19y\right)\times 3=30\times 3$$
- step5: Simplify the equation:
  $$77+13y=90$$
- step6: Move the constant to the right side:
  $$13y=90-77$$
- step7: Subtract the numbers:
  $$13y=13$$
- step8: Divide both sides:
  $$\frac{13y}{13}=\frac{13}{13}$$
- step9: Divide the numbers:
  $$y=1$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{7-4\times 1}{3}$$
- step11: Substitute back:
  $$x=\frac{7-4}{3}$$
- step12: Calculate:
  $$x=1$$
- step13: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=1\end{array}\right.$$
- step14: Check the solution:
  $$\left\{ \begin{array}{l}x=1\y=1\end{array}\right.$$
- step15: Rewrite:
  $$\left(x,y\right) = \left(1,1\right)$$
The solution to the system of equations is $$ (x, y) = (1, 1) $$. This means that one serving of Gerber Multigrain Cereal and one serving of Gerber 2nd Foods Mango satisfies both the calorie and carbohydrate requirements.

### Cost Calculation:
Now, let's calculate the total cost for this combination:
$$
C = 0.30(1) + 0.52(1) = 0.30 + 0.52 = 0.82
$$

### Conclusion:
To provide your child with at least 140 calories and at least 30 grams of carbohydrates at the least cost, you should serve:
- 1 serving of Gerber Multigrain Cereal
- 1 serving of Gerber 2nd Foods Mango

The total cost will be $$ 0.82 $$ (or 82 cents).
To meet the requirements of at least 140 calories and 30 grams of carbohydrates at the least cost, serve 1 serving of Gerber Multigrain Cereal and 1 serving of Gerber 2nd Foods Mango. The total cost will be 82 cents.

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