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Maximum and Minimum Problems help (1 Viewer)

atar90plus

01000101=YES! YES! YES!
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Hi

I need help with these four problems as I spent just about half hour doing these four questions but ended up with the wrong solution. Can you guys please post the full correct working out and solution to these questions. (as attached)

Thanks
 

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tinybrain

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If im correct, these questions are in "Maths in Focus HSC" i remember, back of the book for this chapter, (2)? has full worked out solution.
 

superSAIyan2

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for the surfboard question let the length of the rectangle part be X and the height be 2R ---> thus the radius of the semicircle is R
Perimeter = X + X + 2R + 1/2 (2piR) = 2X + R(pi+2) = 4
From this X = 1/2 [ 4 - R(pi + 2)]

Area of the shape A = 2RX + 1/2 pi R^2
sub X = 1/2 [ 4 - R(pi + 2)] into eqn
A = 4R - R^2(pi+2) + 1/2 pi R^2
dA/dR = 4 - 2R(pi+2) + piR = 4 - R(2pi + 4 - pi) = 4 - R(pi+4)

dA/dR = 0 when R = 4/(pi+4)
PROVE THIS IS A MAX THEN SUB INTO EQN FOR AREA
ANSWER = 1.12m^2
 

superSAIyan2

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and q12 - max area of rectangle occurs when all vertices on circumfgerence - using circle geo it can be shown the the centre of the rectangle (where diagonals intersect ) is the centre of the circle.

Therefore diagonal = diameter = 12cm
let dimensions of rect be xcm by ycm
using pythagoras x^2 +y^2 = 144 ---> y = root(144-x^2)
Area rectangle = xy = x root(144-x^2)

differentiate and you get max area when x = root 72 and y = root 72
therefore max area = 72cm^2
 

atar90plus

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Yes the questions are from the math in focus textbook however they only provide full solutions for mainly the geometry chapters
If im correct, these questions are in "Maths in Focus HSC" i remember, back of the book for this chapter, (2)? has full worked out solution.
 

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