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trapecia [35]
2 years ago
13

1413 A B 1 1 1 2 1 1 1 Each lunch must provide at least 6 units of fat per serving, no more than 7 units of protein, and at leas

t 10 units of carbohydrates. The institution can purchase food A for $0.12 per ounce and food B for $0.08 per ounce. How many ounces of each food should a serving contain to meet the dietary requirements at the least cost
Mathematics
1 answer:
Rainbow [258]2 years ago
6 0

Solution :

It is given that a large institution is preparing food menus which contains foods A as well as B.

Each lunch provides :

10 units of carbohydrates

6 units of fats

7 units of proteins

The food from A cost = $0.12 per ounce

And food from B cost = $0.08 per ounce

Objective function :

$C= 0.12x+0.08y$

Constraints : $x\geq 0, y \geq 0, x+y \geq 6, 2x+y \geq 10, x+y \leq 7$

Domain : $3 \leq x\leq 7$

Range : $0\leq y\leq4$

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5. The recursive algorithm given below can be used to compute gcd(a, b) where a and b are non-negative integer, not both zero.
s2008m [1.1K]

Implementating the given algorithm in python 3, the greatest common divisors of <em>(</em><em>124</em><em> </em><em>and</em><em> </em><em>244</em><em>)</em><em> </em>and <em>(</em><em>4424</em><em> </em><em>and</em><em> </em><em>2111</em><em>)</em><em> </em>are 4 and 1 respectively.

The program implementation is given below and the output of the sample run is attached.

def gcd(a, b):

<em>#initialize</em><em> </em><em>a</em><em> </em><em>function</em><em> </em><em>named</em><em> </em><em>gcd</em><em> </em><em>which</em><em> </em><em>takes</em><em> </em><em>in</em><em> </em><em>two</em><em> </em><em>parameters</em><em> </em>

if a>b:

<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>greater</em><em> </em><em>than</em><em> </em><em>b</em>

return gcd (b, a)

<em>#if</em><em> </em><em>true</em><em> </em><em>interchange</em><em> </em><em>the</em><em> </em><em>Parameters</em><em> </em><em>and</em><em> </em><em>Recall</em><em> </em><em>the</em><em> </em><em>function</em><em> </em>

elif a == 0:

return b

elif a == 1:

return 1

elif((a%2 == 0)and(b%2==0)):

<em>#even</em><em> </em><em>numbers</em><em> </em><em>leave</em><em> </em><em>no</em><em> </em><em>remainder</em><em> </em><em>when</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>2</em><em>,</em><em> </em><em>checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>and</em><em> </em><em>b</em><em> </em><em>are</em><em> </em><em>even</em><em> </em>

return 2 * gcd(a/2, b/2)

elif((a%2 !=0) and (b%2==0)):

<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>odd</em><em> </em><em>and</em><em> </em><em>B</em><em> </em><em>is</em><em> </em><em>even</em><em> </em>

return gcd(a, b/2)

else :

return gcd(a, b-a)

<em>#since</em><em> </em><em>it's</em><em> </em><em>a</em><em> </em><em>recursive</em><em> </em><em>function</em><em>,</em><em> </em><em>it</em><em> </em><em>recalls</em><em> </em><em>the function</em><em> </em><em>with </em><em>new</em><em> </em><em>parameters</em><em> </em><em>until</em><em> </em><em>a</em><em> </em><em>certain</em><em> </em><em>condition</em><em> </em><em>is</em><em> </em><em>satisfied</em><em> </em>

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print()

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print(gcd(4424, 2111))

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