The longest flagpole that could be shipped in a box that measures 2 ft by 2 ft by 12 ft is 2 feet long.
To find the minimum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.
We are given that;
The measurement of box= 2*2*12ft
Now,
To determine the longest flagpole that could be shipped in a box that measures 2 ft by 2 ft by 12 ft, we need to find the maximum length of a flagpole that can fit inside the box without exceeding any of the box's dimensions.
The longest flagpole that could fit inside the box would be equal to the shortest dimension of the box, which is 2 feet. This is because the flagpole would need to fit inside the box in a diagonal orientation in order to maximize its length.
Therefore, by the maxima and minima answer will be 2 feet long.
Learn more about maxima and minima of a function here:
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0.3
0.045 and
0.49
Would really appreciate the help!
Multiply the outside exponent into each
Multiply together
When you multiply an exponent DIRECTLY into another variable(with an exponent), you multiply the exponents.
For example:
(x²)³ =
When you multiply a variable with an exponent into another variable with an exponent, you add the exponents.
For example:
First I multiplied the outside exponents into the numerator and the denominator.
When you have a negative exponent, you move it onto the other side of the fraction to make it positive.
For example:
Sorry if this is confusing