Find two positive numbers satisfying the given requirements. The product is 432 and the sum of the first plus three times the se
cond is a minimum.
1 answer:
Answer:
<h3>36 and 12</h3>
Step-by-step explanation:
Let the two positive integers be x and y.
If their product is 432, then
xy = 432 ......... 1
Also if the sum of the first plus three times the second is a minimum, then;
p(x) = x + 3y
From 1;
y = 432/x ..... 3
Substitute 3 into 2;
p(x) = x+3y
p(x)= x + 3(432/x)
p(x) = x + 1296/x
Since the expression is at minimum when dp(x)/dx = 0
dp/dx = 1 + (-1296)/x²
dp/dx = 1 -1296/x²
0 = 1 -1296/x²
0 = (x²-1296)/x²
cross multiply
0 = x²-1296
x² = 1296
x = √1296
x = 36
Since xy = 432
36y = 432
y = 432/36
y = 12
Hence the two positive numbers are 36 and 12
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<em>hope this helps....</em>