One number is 10 less than twice a second number. Find a pair of such numbers so that their product is as small as possible.
These two numbers are_ and _. (Use a comma to separate your numbers.)
The smallest possible product is_.
The number are NOT 5,15 or 15 and 5
Hmmm.
one number*second number=product.
(2S-10)S=product
2s^2-10s=product.
dp/ds= 4s-10=0
s=2.5 On number then is -5
Let's assume the first number is x and the second number is y.
According to the given information, we can write the equation:
x = 2y - 10
To find the pair of numbers where their product is as small as possible, we need to consider the values that minimize the product xy.
Now, let's express x in terms of y:
x = 2y - 10
Substitute x in terms of y into the equation for the product:
P = xy
P = (2y - 10)y
P = 2y^2 - 10y
To find the minimum product, we can differentiate P with respect to y and set it equal to zero:
dP/dy = 4y - 10 = 0
Solve for y:
4y - 10 = 0
4y = 10
y = 10/4
y = 2.5
Substitute this value of y back into the equation for x to find x:
x = 2(2.5) - 10
x = 5 - 10
x = -5
Therefore, the pair of numbers that yield the smallest possible product is -5 and 2.5, and the smallest possible product is -5 * 2.5 = -12.5. However, since a negative product is not meaningful in this context, the answer would be the next closest numbers that yield the smallest positive product, which are 5 and 2.5.
So, the numbers are 5 and 2.5 and the smallest possible product is 5 * 2.5 = 12.5.
To find a pair of numbers that satisfy the given condition and have the smallest possible product, we can let one of the numbers be x and the other number be y.
Given that one number is 10 less than twice the other number, we can write the equation:
x = 2y - 10
To find the minimum product, we can express the product xy in terms of x alone, substitute the value of x from the equation above, and then identify the minimum value.
The product of the two numbers is given by:
xy = (2y - 10)y
Expanding and simplifying this equation, we get:
xy = 2y^2 - 10y
To find the minimum value of this quadratic expression, we can use calculus by finding the derivative of xy with respect to y and setting it equal to zero:
d(xy) / dy = 0
Differentiating the expression 2y^2 - 10y with respect to y, we get:
2y - 10 = 0
Solving for y, we find:
2y = 10
y = 5
Substituting this value of y back into the equation x = 2y - 10, we can solve for x:
x = 2(5) - 10
x = 10 - 10
x = 0
Therefore, the pair of numbers that satisfy the given condition and have the smallest possible product is 0 and 5. The smallest possible product is 0.