ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: RIIIIII on October 13, 2021, 06:26:13 pm
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(From Cambridge Senior Maths 1/2 Ex 5H)
Assume that angles that look like right angles are right angles.
a i Find an expression for the area A in terms of x and y. (A = (8 + x)y − x^2)
ii Find an expression for the perimeter P in terms of x and y. (P = 2x + 2y + 16)
b i If P = 64 cm, find A in terms of x. (A = 192 + 16x − 2x^2)
ii Find the allowable values for x.
iii Sketch the graph of A against x for these values.
iv What is the maximum area? (224 cm^2)
I can't work out b (ii), the answer for that is 0<x<12
thanks :)
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(From Cambridge Senior Maths 1/2 Ex 5H)
Assume that angles that look like right angles are right angles.
a i Find an expression for the area A in terms of x and y. (A = (8 + x)y − x^2)
ii Find an expression for the perimeter P in terms of x and y. (P = 2x + 2y + 16)
b i If P = 64 cm, find A in terms of x. (A = 192 + 16x − 2x^2)
ii Find the allowable values for x.
iii Sketch the graph of A against x for these values.
iv What is the maximum area? (224 cm^2)
I can't work out b (ii), the answer for that is 0<x<12
thanks :)
Hi,
Never fear! This problem is not as difficult as it seems.
I think there may be a mistake in the answers, but here we go:
We know that x>0, since length can’t be negative. We also know that y>0, meaning that x+8>0, so x>-8. But since that is negative, we don’t count that. Area must be positive, so A>0. This means that you need to solve the quadratic equation for when it is greater than zero. That would give -4√7 + 4 < x < 4√7 + 4. Since x>0, the allowable values of x are 0 < x < 4√7 + 4 (~14.6).
I know you said that answer is 0 < x < 12, but that does not seem mathematically correct. There must be a mistake somewhere in the answers.
Anyway, I hope this helped. All the best!
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Ok using part Bi which you said you solved, you would have found that Y=24-x for a perimeter of 64 right?
And Y can not be below 0 because it is a measurment and a negaitive measurment doens't exist, so x has to be between 0 and 24, but can't be 0 or 24 so its 0<x<24. To be more clear, lets say x=25, then y=-1, which is immposible. Likewise x can't be negative cause its a measurment.
But we can go further, we know that Y is larger than x, so Y-x will be above 0, So sub in Y=24-x into Y-x>0 and you get 24 -x -x>0 or 24-2x>0. This step can be kinda tricky to see but it is just some basic substitution.
Solving for X, we get x<12. So X is above zero and less than 12, which makes sense cause if it was above 12 it would have to be larger than Y lol.
Hope that helps?
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Ok using part Bi which you said you solved, you would have found that Y=24-x for a perimeter of 64 right?
And Y can not be below 0 because it is a measurment and a negaitive measurment doens't exist, so x has to be between 0 and 24, but can't be 0 or 24 so its 0<x<24.
But we can go further, we know that Y is larger than x, so Y-x will be above 0, So sub in Y=24-x into Y-x>0 and you get 24 -x -x>0 or 24-2x>0
Solving for X, we get x<12. So X is above zero and less than 12, which makes sense cause if it was above 12 it would have to be larger than Y lol.
Hope that helps?
Ohh, sorry! Forget what I said, mabajas76 is correct! I don’t know what I was thinking!
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Thank you so much ;D
Ok using part Bi which you said you solved, you would have found that Y=24-x for a perimeter of 64 right?
And Y can not be below 0 because it is a measurment and a negaitive measurment doens't exist, so x has to be between 0 and 24, but can't be 0 or 24 so its 0<x<24. To be more clear, lets say x=25, then y=-1, which is immposible. Likewise x can't be negative cause its a measurment.
But we can go further, we know that Y is larger than x, so Y-x will be above 0, So sub in Y=24-x into Y-x>0 and you get 24 -x -x>0 or 24-2x>0. This step can be kinda tricky to see but it is just some basic substitution.
Solving for X, we get x<12. So X is above zero and less than 12, which makes sense cause if it was above 12 it would have to be larger than Y lol.
Hope that helps?
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Ohh, sorry! Forget what I said, mabajas76 is correct! I don’t know what I was thinking!
That's Ok :) Thank you so much for helping ;D
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That's Ok :) Thank you so much for helping ;D
No worries :)