Answer:
125 and 185
Step-by-step explanation:
first u have to use the table values and put them in the linear equation.
so you have in first: w= 7, so: h= 20.7-15-->h=125
and in the last: w=10, so h=20.10-15-->h=185
Given : ABCF is a Rectangle
⇒ AB = CF
Given : AB = 6 cm
⇒ CF = 6 cm
Perimeter of a Rectangle is given by : 2(Length + Width)
Given : Perimeter of the Rectangle = 20 cm
⇒ 2(6 + BC) = 20
⇒ 6 + BC = 10
⇒ BC = 4
Given : BD = 9
But, We can See that : BD = BC + CD
⇒ CD + 4 = 9
⇒ CD = 5
We know that :
✿ 
Where : a and b are Lengths of Parallel Sides and Height is Length between those Parallel Sides
In the Given Trapezium :
The Parallel Sides are CD and EF and Height is CF
Given : EF = 3 cm
288 divided by 12 = 24
240 divided by 10 = 24
Answer:A
Step-by-step explanation:A
The questions for this problem would be:
1. What is the dimensions of the box that has the maximum volume?
2. What is the maximum volume of the box?
Volume of a rectangular box = length x width x height
From the problem statement,
length = 12 - 2x
width = 9 - 2x
height = x
where x is the height of the box or the side of the equal squares from each corner and turning up the sides
V = (12-2x) (9-2x) (x)
V = (12 - 2x) (9x - 2x^2)
V = 108x - 24x^2 -18x^2 + 4x^3
V = 4x^3 - 42x^2 + 108x
To maximize the volume, we differentiate the expression of the volume and equate it to zero.
V = 4x^3 - 42x^2 + 108x
dV/dx = 12x^2 - 84x + 108
12x^2 - 84x + 108 = 0x^2 - 7x + 9 = 0
Solving for x,
x1 = 5.30 ; Volume = -11.872 (cannot be negative)
x2 = 1.70 ; Volume = 81.872
So, the answers are as follows:
1. What is the dimensions of the box that has the maximum volume?
length = 12 - 2x = 8.60
width = 9 - 2x = 5.60
height = x = 1.70
2. What is the maximum volume of the box?
Volume = 81.872
