6. Sketch the region enclosed by the graphs of x =0, 6y-5x=0 and x+3y=21. Find the area
2 answers:
We can see that revolving the region formed by intersecting 3 lines, we will get 2 cones that are connected their bases.
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50
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x = side length of the base.
Volume of the box is
.
So
Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.
![4x = \frac{460}{ x^{2} } \\ 4x^{3} = 460 \\ x^{3} = 115 \\ x = \sqrt[3]{115} = 4.86](https://tex.z-dn.net/?f=%204x%20%3D%20%5Cfrac%7B460%7D%7B%20x%5E%7B2%7D%20%7D%20%20%5C%5C%20%204x%5E%7B3%7D%20%3D%20460%20%20%5C%5C%20x%5E%7B3%7D%20%3D%20115%20%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B115%7D%20%3D%204.86%20)
Then
So the box is 4.86 in. wide and 4.87 in. high.
x = side length of the base.
Volume of the box is
So
Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.
Then
So the box is 4.86 in. wide and 4.87 in. high.