Answer:
Length = 9 inches
Step-by-step explanation:
The perimeter of a rectangle is:
perimeter = 2(length + width)
24 = 2(a+b)
a = b + 6
a = length
b = width
then:
24 = 2((b+6)+b)
24/2 = b+6+b
12 = 2b + 6
12-6 = 2b
6 = 2b
b = 6/2
b = 3 inches
a = b+6
a = 3+6
a = 9 inches
Check:
24 = 2(9+3)
24 = 2*12
Answer:
the area of the problem is 20
The radius of tire is larger than radius of wheel by 5 inch
Step-by-step explanation:
We know that circumference of the circle is 2πr where “r” is the radius of the circle
Circumference refers to the dimension of the periphery of the circle. Since tires are put on the periphery of the wheel hence, we considered the circumferential aspect of the wheel.
Given-
Circumference of tires= 28π inches
2πr= 28π cancelling the common term “π” both sides
r (radius of the tires) = 14 inches
Circumference of the wheel rims= 18π
2πr= 18π cancelling the common term “π” both sides
r (radius of the tires) = 9 inches
Difference between the radius= 14-9= 5 inches
Hence, the difference between the radius of tires and the radius of the wheels is 5 inches
The dimensions of a box that have the minium surface area for a given Volume is such that it is a cube. This is the three dimensions are equal:
V = x*y*z , x=y=z => V = x^3, that will let you solve for x,
x = ∛(V) = ∛(250cm^3) = 6.30 cm.
Answer: 6.30 cm * 6.30cm * 6.30cm. This is a cube of side 6.30cm.
The demonstration of that the shape the minimize the volume of a box is cubic (all the dimensions equal) corresponds to a higher level (multivariable calculus).
I guess it is not the intention of the problem that you prove or even know how to prove it (unless you are taking an advanced course).
Nevertheless, the way to do it is starting by stating the equations for surface and apply two variable derivation to optimize (minimize) the surface.
You do not need to follow with next part if you do not need to understand how to show that the cube is the shape that minimize the surface.
If you call x, y, z the three dimensions, the surface is:
S = 2xy + 2xz + 2yz (two faces xy, two faces xz and two faces yz).
Now use the Volumen formula to eliminate one variable, let's say z:
V = x*y*z => z = V /(x*y)
=> S = 2xy + 2x [V/(xy)[ + 2y[V/(xy)] = 2xy + 2V/y + 2V/x
Now find dS, which needs the use of partial derivatives. It drives to:
dS = [2y - 2V/(x^2)] dx + [2x - 2V/(y^2) ] dy = 0
By the properties of the total diferentiation you have that:
2y - 2V/(x^2) = 0 and 2x - 2V/(y^2) = 0
2y - 2V/(x^2) = 0 => V = y*x^2
2x - 2V/(y^2) = 0 => V = x*y^2
=> y*x^2 = x*y^2 => y*x^2 - x*y^2 = xy (x - y) = 0 => x = y
Now that you have shown that x = y.
You can rewrite the equation for S and derive it again:
S = 2xy + 2V/y + 2V/x, x = y => S = 2x^2 + 2V/x + 2V/x = 2x^2 + 4V/x
Now find S'
S' = 4x - 4V/(x^2) = 0 => V/(x^2) = x => V =x^3.
Which is the proof that the box is cubic.
