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Tyace needs at least 40 hot dogs and 40 buns for a cookout. Packages of hotdogs cost $3.50 and contains 10 hotdogs. Packages of buns cost$2.50 and contains 8 buns. The maximum amount of money Tyace can spend on hot dogs and buns is $70. What are possible combinations of packages that Tyace can buy? Jan 29, 2020 ### 2+0 Answers #1 +24388 +2 Tyace needs at least 40 hot dogs and 40 buns for a cookout. Packages of hotdogs cost$3.50 and contains 10 hotdogs.
Packages of buns cost $2.50 and contains 8 buns. The maximum amount of money Tyace can spend on hot dogs and buns is$70.
What are possible combinations of packages that Tyace can buy?

I assume:

$$\text{Let packages of hotdogs =x and x_{min} = 4 packages } \\ \text{Let packages of buns =y and y_{min} = 5 packages }$$

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{3.50}{10~hotdogs} * 10x + \dfrac{2.50}{8~buns} * 8y} &=& \mathbf{70} \\\\ \dfrac{3.50}{ hotdogs} * x + \dfrac{2.50}{ buns} * y &=&  70 \\\\ 3.50x + 2.50y &=& 70 \quad | \quad * 2 \\\\ \mathbf{7x + 5y} &=& \mathbf{140} \\ \hline \end{array}$$

Euler :

$$\begin{array}{|rclrclrcl|} \hline \mathbf{7x + 5y} &=& \mathbf{140} \\ 5y &=& 140 - 7x \\ y &=& \dfrac{140 - 7x}{5} \\ y &=& \dfrac{140 - 5x-2x}{5} \\ y &=& 28 - x - \underbrace{\dfrac{2x}{5}}_{=a} \\ y &=& 28 - x - a & a&=&\dfrac{2x}{5} \\ & & & 5a&=&2x \\ & & & 2x&=&5a \\ & & & x&=&\dfrac{5a}{2} \\ & & & x&=&\dfrac{4a+a}{2} \\ & & & x&=&2a+\underbrace{\dfrac{a}{2}}_{=b} \\ & & & x&=&2a+b & b&=& \dfrac{a}{2} \\ & & & & & & 2b&=& a \\ & & & & & & \mathbf{a}&=& \mathbf{2b} \\ & & & x&=&2(2b)+b \\ & & & \mathbf{x}&=&\mathbf{5b} \\ y &=& 28 - 5b - 2b \\ \mathbf{y} &=& \mathbf{28 - 7b} \\ \hline \end{array}$$

The possible combinations of packages:

$$\begin{array}{|c|r|r|} \hline b & x=5b & y = 28 - 7b \\ \hline 1 & 5 & 21 \\ 2 & 10 & 14 \\ 3 & 15 & 7 \\ \hline \end{array}$$

1.   5 packages of hotdogs and 21 packages of buns

2. 10 packages of hotdogs and 14 packages of buns

3. 15 packages of hotdogs and 7 packages of buns

Jan 29, 2020
#2
+21931
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He needs 40 buns and 40 dogs ...so he HAS to spend

40 dogs/10dogs/pkg   x   $3.50/pkg =$14       (4 packages dogs)

40 buns/8buns/pkg     x    $2.50/pkg = 12.50 (5 packages buns) 14 + 12.50 = 26.50 He has 70 - 26.50 = 43.50 left to spend Now any combo that spends 43.50 is allowed as he has met his 40 dog-bun requirement 3.50 y + 2.50 x = 43.50 I used a graphical approach to see where x,y pairs are integer values (since you cannot split packages) you will see that x, y = 2,11 9,6 and 16, 1 all fit the bill ....remember you have to add the obligatory red packages above ! y is dogs x is buns so 7 b 15 d 14 b 10 d and 21 b 5 d all would spend the total$70

Jan 29, 2020
edited by ElectricPavlov  Jan 29, 2020