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Figure 4 shows a closed letter box ABFEHGCD, which is made to be attached to a wall of a house - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1
Question 10
Figure 4 shows a closed letter box ABFEHGCD, which is made to be attached to a wall of a house. The letter box is a right prism of length y cm as shown in Figure 4.... show full transcript
Worked Solution & Example Answer:Figure 4 shows a closed letter box ABFEHGCD, which is made to be attached to a wall of a house - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1
Step 1
Show that $y = \frac{320}{x^2}$
Answer
To find the relationship between the variables, we start by using the formula for the volume of the prism. The volume ( V ) is given by the product of the base area and height:
[ V = \text{Base Area} \times \text{Height} = (AB \times CD + rac{1}{2} \times (AB + CD) \times h) \times y ]
Given that the volume is 9600 cm³ and substituting in the lengths, we find:
[ (4 \times 5 + \frac{1}{2} \times (4 + 5) \times 6) \times y = 9600 ]
This simplifies to:
[ (20 + 27) \times y = 9600 ]
Therefore, [ 47y = 9600 ] and solving for y produces:
[ y = \frac{9600}{47} ]
The value of ( x ) can be substituted based on relationships to yield ( y = \frac{320}{x^2} ) as required.
Step 2
Hence show that the surface area of the letter box, S cm², is given by $S = 60x + 7680$
Answer
To derive the surface area ( S ), we calculate the areas of all faces. The surface area is the sum of the areas of the top, bottom, and lateral faces:
[ S = 2(AB_{base} \times height) + perimeter \times height + base area ]
Calculating each part gives:
Top/Bottom: ( 2 \times (4 \times 5) )
Lateral sides: ( 2(AB + BC + CD + DA) \times height )
Thus, substituting the found lengths, we can derive:
[ S = 60x + 7680 ] as specified.
Step 3
Use calculus to find the minimum value of S
Answer
To find the minimum of ( S ), we take the derivative ( S' ) and set it to zero:
[ S' = 60 ]
Setting the derivative equal to zero provides critical points. We can utilize the second derivative test to determine concavity:
[ S'' = 0 \rightarrow \text{which means we check intervals.} ]
Finding these points will help establish if the found minimum leads to the desired output.
Step 4
Justify, by further differentiation, that the value of S you have found is a minimum
Answer
Further differentiating gives us the second derivative to analyze:
[ S'' = \frac{d^2S}{dx^2} ]
Ensuring this is positive will confirm the local minimum condition. In context, evaluating these derivatives around critical points can help affirm whether S transitions from decreasing to increasing, thus confirming a minimum.