From a square piece of cardboard paper of area size 9 m2 , squares of the same size are cut off from each corner of the paper. this allows for folding the paper to a box without a top. what values for the lengths of the squares only make sense? how do we have to choose the length of the squares to maximize the volume of the box? verify that your answer is indeed a maximum.

Max vol = 2 cubic metres

Step-by-step explanation:

Given that from a square piece of cardboard paper of area size 9 m2 , squares of the same size are cut off from each corner of the paper.

Side of the square = 3m

If squares are to be cut from the corners of the cardboard we have the dimensions of the box as

3-2x, 3-2x and x.

Hence x can never be greater than or equal to 1.5

V(x) = Volume =

We use derivative test to find the maxima

Equate I derivative to 0

If x= 3/2 box will have 0 volume

So this is ignored

V"(1/2) <0

So maximum when x =1/2

Maximum volume

= cubic metres

answer

explanation

the area of a circle is calculated using the formula

or

we can use any of the above formulae to calculate the required. preferably, let us use the latter since the diameter is

we substitute the diameter into the formula to get:

this is the exact value for the area of the circle.

the correct answer is a

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.