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mindofnyny
13.07.2019 •
Mathematics
Find two positive numbers such that their product is 192 and their sum is a minimum without calculus
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Ответ:
192 = 2^2*2*2*2*2*3
= 16 * 12
= 6 * 32
= 64 * 3
= 8 * 24
= 4 * 48
Looks like its 16 and 12
Ответ:
Since we want to solve for the variable x, we want to isolate x
a²x + (a - 1) = (a + 1)x ⇒ Distribute x to (a+1). Also, remove parentheses
a²x + a - 1 = ax + x ⇒ Subtract a from both sides
a²x - 1 = ax + x - a ⇒ Add 1 to both sides
a²x = ax + x - a + 1 ⇒ Subtract (ax + x) from both sides
a²x - (ax + x)= ax + x - a + 1 - (ax+x) ⇒ Simplify. Remember that multiplying positive by negative = negative
a²x - ax - x = ax + x - a + 1 - ax - x ⇒ Simplify
a²x - ax - x = -a + 1 ⇒ Factor out the x from a²x - ax - x
x(a² - a - 1) = -a + 1 ⇒ Divide both sides by (a² - a - 1)
x = (-a + 1) / (a² - a - 1)
However, we need to make sure that the denominator does not equal 0. Therefore, you set the denominator = 0 (just use the quadratic formula for this), and it gives that the denominator =0 when a = (1+√5)/2 AND (1-√5)/2
Therefore, the final answer is
x = (-a + 1) / (a² - a - 1) given that a ≠ (1+√5)/2, a ≠ (1-√5)/2