Question

the numbers p,q,r and s are positive integer satisfy the equation p+2q+3r+4s=k 4p=3w=2r=s choose p,q,r and s so that the value of k is minimum then the value of pq+rs is
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Answer 1

Answer:

Step-by-step explanation:

Given p,q, r and s are positive integers and

4p = 3q = 2r = s -- eq.1

Since s is an integer, it must be a multiple of 4, 3 and 2

LCM(4,3,2) = 12

Substitute s = 12

By equating equation 1, we get

p = 3

q = 4

r = 6

pq + rs = 3*4 + 6*12 = 12 + 72 = 84

The answer in answer key is 72 but i'm not sure about it.

KE_LT - 22

Answer 2

Given :   numbers p,q,r and s are positive integer satisfy the equation p+2q+3r+4s=k   , 4p=3w=2r=s  

To find : value of pq+rs

Solution:

p+2q+3r+4s=k  

4p=3q=2r=s

Find LCM of 4 , 3 , 2   = 12

=> 4p=3q=2r=s = 12

=> p = 3

    q = 4

     r  = 6

    s  = 12

p+2q+3r+4s=k  

=> 3  + 2*4  + 3*6  + 4*12 = k

=> 3 + 8  + 18 + 48  = k

=> k = 77

pq+rs   =  (3)(4)  + (6)(12)

= 12  + 72

= 84

value of pq+rs = 84

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