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: t=3

step-by-step explanation:

step-by-step explanation:

let's assume

number of candles set =x

price of of x candles =p(x)

we are given

a candle maker prices one set of scented candles at $10 and sells an average of 200 sets each week

so, we get point as (200,10)

he finds that when he reduces the price by $1, he then sells 50 more candle sets each week

so, we can find slope

now, we can use point slope form of line

now, we can plug values

we can use point

and find b

now, we can plug it back

and we get

now, we can find revenue

we know that

revenue = price*quantity

so, we get