Minimum and Maximum
- Thread starter lem0n84
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How many small loaves of bread will be sold for $1035?
How many large loaves of bread will be sold for $1035?
Please show us what you have tried and exactly where you are stuck.
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Both small and large breads were sold together. And I need to find the minimum number of breads (which has both small and large) sold when the total money made is $1065. And then I need to find the maximum number of breads sold when the total money raised is $1065.
I thought of the equation 25x + 35 y = 1065.
But I don't know how to go on solving it.
Final answer for minimum number of breads sold is 31, and the maximum number of breads sold is 41.
For the minimum number of breads sold,
By trial and error, I did find
$25 x 29 small loaf + $35 x 2 large bread = 1065.
But I need to prove this mathematically.
Help please?
I thought of the equation 25x + 35 y = 1065.
But I don't know how to go on solving it.
Final answer for minimum number of breads sold is 31, and the maximum number of breads sold is 41.
For the minimum number of breads sold,
By trial and error, I did find
$25 x 29 small loaf + $35 x 2 large bread = 1065.
But I need to prove this mathematically.
Help please?
Dr.Peterson
Elite Member
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- Nov 12, 2017
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- 16,087
If you are intended to use only arithmetic, then trial and error may be the best way.
If you know anything about Diophantine equations, then find the general non-negative integer solution to 25x + 35 y = 1065, and find the pairs that yield the largest and smallest value for x+y.
Please tell us the context of the problem, and what you have learned.
You can solve this using Diophantine equations, but you do not need that much fire power.
[math]41 < 1035 \div 25 < 42.[/math]
[math]1035 - (41 * 25) = 10 \implies 1035 - (40 *25) = 35.[/math]
Maximum number is 41, 40 small and 1 large.
Now you figure out the minimum.
[math]41 < 1035 \div 25 < 42.[/math]
[math]1035 - (41 * 25) = 10 \implies 1035 - (40 *25) = 35.[/math]
Maximum number is 41, 40 small and 1 large.
Now you figure out the minimum.
becoolbe777
New member
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- Feb 26, 2022
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25x+35y=1065
X min 2 and y 29
Total 31 min
Y min 4 x 37
Total max 41 max
X min 2;y =29 total 31 min
Y min 4 ;x =37 total 41 max
X min 2 and y 29
Total 31 min
Y min 4 x 37
Total max 41 max
X min 2;y =29 total 31 min
Y min 4 ;x =37 total 41 max
Let x be the number of small loaves sold and let y be the number of large loaves sold
25x+ 35y= 1065. Simplify by dividing through by 5:
5x+ 7y= 213
Since x and y are numbers of things sold, they must be non-negative integers. This is a "Diophantine equation".
Start by looking at 5x+ 7y= 1, 5 divides into 7 once with remainder 2: 7- 5= 2. 2 divides into 5 twice with remainder 1: 5- 2(2)= 1. 5- 2(7- 5)= 3(5)- 2(7)= 1. Now multiply by 213: 639(5)- 426(7)= 213.
One solution to 5x+ 7y= 213 is x= 639, y= -426. That is not a valid answer because y is negative.
But x= 639- 7n and y= -426+ 5n is also a solution for any n because 5(639- 7n)+ 7(-426+ 5n)= (639(5)- 425(7))+ (-35n+ 35n)= 213.
In order that y be non-negative, n must be at least 426/5= 85.2 so take n= 86.
x= 639- 7(86)= 639- 602= 37 and y= -426+ 5(86)= -426+ 430= 4.
There could have been 37 small loaves and 4 large loaves sold, a total of 41 loaves: 25(37)+ 35(4)= 925+ 140= $1065.
In order that x be non-negative, n must be less than 639/7= 91.28... so take n= 91.
x= 639- 7(91)= 639- 637= 2 and y= -426+ 5(91)= -426+ 455= 29.
There could have been 2 small loaves and 29 large loaves, a total of 31 loaves; 25(2)+ 35(29)= 50+ 1015= $1065.
The least number of loaves sold 31, 2 small and 29 large loaves. the most that could have been sold was 41, 37 small and 4 large loaves.