Question

Diagram 3 shows a piece of rectangularcardboard and an open box that is made from the cardboard.The box is made by cutting out four squares of equal size from the cornersof the cardboard then folding up the sides. Finda) the length in cm of sides of the squares to be cut out in order to get a box with largest volume.b) the minimum number of the boxes needed to fill with 5645 cm³ of pudding

Answer 1

Step 1:

In this question, we are given the following:

Diagram 3 shows a piece of rectangular cardboard and an open box that is made from the cardboard.



The box is made by cutting out four squares of equal size from the corners

of the cardboard then folding up the sides.

Find



a) the length in cm of sides of the squares to be cut out in order to get a box with the largest volume.

`\begin{gathered} The\text{ volume of the rectangle would be expressed as:} \\ \text{V = ( 30-2x )(16-2x) ( x)} \\ Multiply\in g\text{ out, we have that:} \\ V=480x-92x^2+4x^3 \\ \text{Differentiating V with respect to x, we have that:} \\ (dV)/(dx)=480-184x+12x^2=0 \\ \text{Factorizing the quadratic equation, we have that:} \\ \text{x = 12 or x =}(10)/(3) \end{gathered}`
`\begin{gathered} \text{Differentiating again, we have that:} \\ \frac{d^2V}{dx^2\text{ }}\text{ = -184 + 24 x} \end{gathered}`

To get the maximum, we need to substitute the values of :

`\begin{gathered} x\text{ = 12, we have that:} \\ \frac{d^2V^{}}{dx^2\text{ }}\text{ = -184 + 24( 12) = }-184\text{ +288 = 104} \\ x=(10)/(3),\text{ we have that:} \\ (d^2V)/(dx^2)\text{ = -184 + 24 (}(10)/(3))\text{ = -184 +}(240)/(3)\text{ = - 184 + 80 = -104 }<0 \end{gathered}`

At this stage, we can see that:

`\begin{gathered} x\text{ =}\frac{10\text{ }}{3}cm\text{ is the length of the squares to be cut in order to get a box with } \\ \text{largest volume} \end{gathered}`

b) Find the minimum number of the boxes needed to fill with 5645 cm³ of pudding​

`\begin{gathered} \text{From the equation,} \\ V=(30-2\text{x )(16-2x)(x)} \\ \text{put x =}(10)/(3)\text{ in the equation, we have that:} \\ V\text{ = \lbrack}30\text{ -2(}(10)/(3))\rbrack\text{ \lbrack 16-2(}(10)/(3)\rbrack\lbrack(10)/(3)\rbrack \\ V\text{ = ( 30 -}(20)/(3))\text{ ( 16 - }(20)/(3))((10)/(3)) \\ V=725.93cm^3 \\ Now\text{, we asked to find the minimum number of boxes ne}eded^{} \\ to^{} \\ \text{fill with 5645cm}^{3\text{ }}\text{ of pudding.} \\ \text{Then, we ne}ed\text{ to do the following:} \end{gathered}`

Minimum number of boxes =

`\begin{gathered} (5645)/(725.93) \\ =\text{ 7.78} \\ \approx\text{ 8} \end{gathered}`

CONCLUSION:

A minimum of 8 boxes will be needed to fill with 5645 cm³ of pudding​