Calculus - Differentiation - Applied max/min questions.
Type 2: 2D shapes - Test Yourself 1 - Solutions.
Creating 2 shapes | 1. (i)
(ii) |
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2. | |||||||||
Areas and paddocks. | 3. (i) (ii) |
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4. Let the dimensions of the printed area be x cm wide and y cm high. Hence area of printed material is A = xy = 280 cm2. |
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5.
(i) PA was 12 cm long. After the fold, x cm (KA) remains vertical so the balance KP1 must be the difference - so (12 - x) cm. (ii) (iii) |
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6. (i) PD = 10 - x = AQ = BR = SC. So we can apply Pythagoras' Theorem to calculate (say) PS. PS2 = x2 + (10-x)2. We could take the square root to calculate PS but, So the area of the square PQRS is A = x2 + (10-x)2 |
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Geometric shapes. | 7.
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8. (i)
(ii) |
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9. (i)
(ii) (iii)
So change of gradient around x = 10 from -ve to +ve - hence a maximum value for A. ∴ BC = 10 cm. |
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10. (i) (ii) (iii)
Change of gradient -ve to +ve so a maximum value. |
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11. (i) ABCD is a rectangle and so AB//DCF and EAD//BC. (ii) Ratioing corresponding sides in silimar triangles: (iii) ED = x + 4 and DF = 6 + y (iv)
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Distance between 2 curves. |
12. (i)
(ii) D = top curve - bottom curve.
(iii) |
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13. |