Kelly2713:
Good afternoon my able maths gurus. I love this forum so much in that i have learn alot from you guys and also appreciate ur efforts and contributions to this great forum.

So, my kid brough this to me and i was like., is this girl trying to crack my brain?

A sheet metal is used for making a closed rectangular box with square ends. If the total area of the metal is 216cm², find the dimension that will give the greatest volume.




2. From the result of the soil test carried out on the school farm ear marked for cowpea production, it was discovered that 37kg of phosphorus per heitare will be needed to ensure good yield. If single superphosphate fertilizer is to use as the source of phosphorus. Calculate the amount or weight of single superphosphate fertilizer that will be required to supply the amount of phosphorus required. (single superphosphate fertilizer contains 18% phosphorus, a bag is 100kg).

Thanks

sparog:


This one na maths or soil chemistry?

FromZeroToHero:
how much will you pay?

Kolombz:
Question 1 involves calculus:
The box consists of 4 rectangles and 2 squares.From this info,we can work out the formula for the surface area of the box:
The surface area of a rectangle is:xh.since there are 4 rectangles,the total surface area would be 4xh.
For the 2 squares,the total area is 2x^2.
Therefore the total surface area for the box is:
4xh+2x^2=216.......[1]
volume of box is x^2h......[2]
Make h the subject of formula in equation 1 and substitute in equation 2.
h=216-2x^2/4x.......[3]
Therefore V=x^2(216-2x^2/4x)=54x-(x^3/2)....(4)on simplification.
According to rule of calculus,The volume is maximum when dV/dx=0
on differentiating equation 4,
We get dV/dx=54-(3x^2/2)=0.....(5)
Solve for x in equation 5
54=3x^2/2
3x^2=108
X^2=36
x=6
Now we have gotten x,substitute its value in equation 3 to get h:h=216-2(6)^2/4(6)
h=6
So that dimension that can produce a box of maximum volume is(6,6) I.e Two squares with a length of 6 cm each and 4 rectangles with a length of 6cm and a breadth of 6cm each are what is required for a box of maximum volume.

Thanks bro

Kolombz:
Question 1 involves calculus:
The box consists of 4 rectangles and 2 squares.From this info,we can work out the formula for the surface area of the box:
The surface area of a rectangle is:xh.since there are 4 rectangles,the total surface area would be 4xh.
For the 2 squares,the total area is 2x^2.
Therefore the total surface area for the box is:
4xh+2x^2=216.......[1]
volume of box is x^2h......[2]
Make h the subject of formula in equation 1 and substitute in equation 2.
h=216-2x^2/4x.......[3]
Therefore V=x^2(216-2x^2/4x)=54x-(x^3/2)....(4)on simplification.
According to rule of calculus,The volume is maximum when dV/dx=0
on differentiating equation 4,
We get dV/dx=54-(3x^2/2)=0.....(5)
Solve for x in equation 5
54=3x^2/2
3x^2=108
X^2=36
x=6
Now we have gotten x,substitute its value in equation 3 to get h:h=216-2(6)^2/4(6)
h=6
So that dimension that can produce a box of maximum volume is(6,6) I.e Two squares with a length of 6 cm each and 4 rectangles with a length of 6cm and a breadth of 6cm each are what is required for a box of maximum volume.

Kylekent59:



He made a mistake at the Bolded


h= 216/4x - 2x^2/4.


Better inform the guy

Kylekent59:
cool

ikbnice:
What class is your kid if I may ask

ikbnice:
What class is your kid if I may ask

sparog:


This one na maths or soil chemistry?

Kolombz:
Question 1 involves calculus:
The box consists of 4 rectangles and 2 squares.From this info,we can work out the formula for the surface area of the box:
The surface area of a rectangle is:xh.since there are 4 rectangles,the total surface area would be 4xh.
For the 2 squares,the total area is 2x^2.
Therefore the total surface area for the box is:
4xh+2x^2=216.......[1]
volume of box is x^2h......[2]
Make h the subject of formula in equation 1 and substitute in equation 2.
h=216-2x^2/4x.......[3]
Therefore V=x^2(216-2x^2/4x)=54x-(x^3/2)....(4)on simplification.
According to rule of calculus,The volume is maximum when dV/dx=0
on differentiating equation 4,
We get dV/dx=54-(3x^2/2)=0.....(5)
Solve for x in equation 5
54=3x^2/2
3x^2=108
X^2=36
x=6
Now we have gotten x,substitute its value in equation 3 to get h:h=216-2(6)^2/4(6)
h=6
So the dimension that can produce a box of maximum volume is(6,6) I.e Two squares with a length of 6 cm each and 4 rectangles with a length of 6cm and a breadth of 6cm each are what is required for a box of maximum volume.

Kelly2713:
Good afternoon my able maths gurus. I love this forum so much in that i have learn alot from you guys and also appreciate ur efforts and contributions to this great forum.

So, my kid brough this to me and i was like., is this girl trying to crack my brain?

A sheet metal is used for making a closed rectangular box with square ends. If the total area of the metal is 216cm², find the dimension that will give the greatest volume.




2. From the result of the soil test carried out on the school farm ear marked for cowpea production, it was discovered that 37kg of phosphorus per heitare will be needed to ensure good yield. If single superphosphate fertilizer is to use as the source of phosphorus. Calculate the amount or weight of single superphosphate fertilizer that will be required to supply the amount of phosphorus required. (single superphosphate fertilizer contains 18% phosphorus, a bag is 100kg).

Thanks

Kelly2713:
Good afternoon my able maths gurus. I love this forum so much in that i have learn alot from you guys and also appreciate ur efforts and contributions to this great forum.

So, my kid brough this to me and i was like., is this girl trying to crack my brain?

A sheet metal is used for making a closed rectangular box with square ends. If the total area of the metal is 216cm², find the dimension that will give the greatest volume.

Thanks

Martinez39:

1) The total area of the metal sheet is equal to the total area of the box. Given that the box has square ends and rectangular sides the the surface area SA of the closed box is SA = 4yx + 2x² = 216 , this reduces to 2yx + x² = 108 .

2) The volume V of the box is given by V = V(x,y) = x²y . Solving for y in the surface area gives y = (108 - x²) ÷ 2x . Substituting this into the volume equation yields V = V(x) = 54x - (½)x³ . This is the equation to be maximised and by logic, the domain of x is the open interval 0 < x < (216)^½.

3) To find the value of x that yields the maximum volume, you need to find the critical number(s) of the volume function. The critical number for this equation will only occur at x - values for which V = V'(x) = 0 . Given that dV/dx = V'(x) = 54 - 2(¾)x² , the x - values for which dV/dx = V'(x) = 0 are ±6 , we pick 6 because it lies in the feasible domain of x . So the value of x that maximises V = V(x) is 6cm.

4) to find the corresponding value of y that maximises V = V(x) = V(x,y) , put the value of x that maximises V= V(x) into the surface area equation and solve for y . Doing this yields y = 6cm .

5) So if the square ends of both boxes are 6cm × 6cm and the four rectangular sides are of the dimensions 6cm × 6cm , you will get the maximum volume.