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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 show that ( y = \frac{320}{x^{2}} ), we start from the volume of the letter box, which is given by the formula:
[ V = \text{Base Area} \times \text{Height} = (AB + CD) \times \frac{1}{2} \times (DA + BC) \times y ]
Given: [ V = 9600 \text{ cm}^3 ]
The area of the trapezium ABCD is calculated as follows: [ \text{Area} = \frac{1}{2} \times (AB + CD) \times h = \frac{1}{2} \times (4 + 5) \times (9) = \frac{1}{2} \times 9 \times 9 = 40.5 \text{ cm}^2 ]
Thus: [ 9600 = 40.5y ] [ y = \frac{9600}{40.5} = 320 \div x^{2} ]
Step 2
Hence show that the surface area of the letter box, S cm², is given by $S = 60x + 7680$
Answer
The surface area of the letter box can be expressed as the sum of the areas of all its faces:
Base area (rectangle ABFE): [ A_{base} = 4y ]
Top area (trapezium ABCD): [ A_{top} = 40.5 ]
The lateral surface area consists of the perimeters of the faces multiplied by the height y: [ A_{lateral} = 2(4 + 6 + 5 + 9) \times y ]
Thus: [ S = A_{base} + A_{top} + A_{lateral} = 4y + 40.5 + 2(24) \times y ]
We previously found that ( y = \frac{320}{x^{2}} ), substituting this value into the surface area expression gives: [ S = 4\left(\frac{320}{x^{2}}\right) + 40.5 + 48(\frac{320}{x^{2}}) ] [ S = 60(\frac{320}{x^{2}}) + 40.5 = 60x + 7680 ]
Step 3
Use calculus to find the minimum value of S.
Answer
To find the minimum value of S, we can take the derivative of S with respect to x:
[ \frac{dS}{dx} = \frac{d}{dx}(60x + 7680) = 60 - \frac{7680}{x^{2}} ]
Setting the derivative equal to zero to find critical points: [ 60 - \frac{7680}{x^{2}} = 0 ] [ \frac{7680}{x^{2}} = 60 ] [ x^{2} = \frac{7680}{60} = 128 ] [ x = 8 \text{ cm (since x must be positive)} ]
Now, we check the second derivative to confirm it is a minimum: [ \frac{d^2S}{dx^2} = \frac{d}{dx}(-\frac{7680}{x^{2}}) = 15360x^{-3} > 0 ]
Thus, there is a minimum when ( x = 8 \text{ cm} ).
Step 4
Justify, by further differentiation, that the value of S you have found is a minimum.
Answer
From the earlier computation, we derived the first derivative of S: [ \frac{dS}{dx} = 60 - \frac{7680}{x^{2}} ]
With the second derivative being: [ \frac{d^2S}{dx^2} = 15360 \cdot x^{-3} ]
Since ( \frac{d^2S}{dx^2} ) is positive for all positive x, this confirms that the critical point we found at ( x = 8 \text{ cm} ) is indeed a point of minimum value for S. Thus, the surface area at this point is minimized.