# C2 questionWatch

#1
i really cant do this question .
its too long to type it up . so ive uploaded it.

(dont worry its not a virus)
0
9 years ago
#2
(Original post by awargi)
i really cant do this question .
its too long to type it up . so ive uploaded it.

(dont worry its not a virus)
a) P = 80 = 2x + 2y + 2px/2

Rearrange to get y=... and away you go.

b) Differentiate to get 80-(4+p)x, put this equal to 0 to get your stationary value of x.

c) Find the second derivative (differentiate the first derivative again) and then put your answer to b) into it. If it is a maximum, the answer should be less than zero.

d) Put the maximum value of x that you found in b) back into the original equation.
0
9 years ago
#3
(a) Get the rectangle and semicircle area separately, and add them together. (That is, length*height, and .)

(b) Differentiate the equation that they've given you in part (a), then solve for to determine value of x at the stationary point.

(c) Take the second derivative of the the equation they gave you in part (a). Find the value of the second derivative for the value of x that you found in part (b). If this value is negative, then the stationary point is a maximum.

(d) Use the equation they gave you in part (a), with the value of x that you found in part (b). That will enable you to calculate A.

I hope that made some sense.
0
#4
(Original post by fishpie57)
a) P = 80 = 2x + 2y + 2px/2

Rearrange to get y=... and away you go.

b) Differentiate to get 80-(4+p)x, put this equal to 0 to get your stationary value of x.

c) Find the second derivative (differentiate the first derivative again) and then put your answer to b) into it. If it is a maximum, the answer should be less than zero.

d) Put the maximum value of x that you found in b) back into the original equation.

thnx alot
but the method for part (a) is some next method. lol
0
9 years ago
#5
(Original post by sa55afras)
(a) Get the rectangle and semicircle area separately, and add them together. (That is, length*height, and .)

(b) Differentiate the equation that they've given you in part (a), then solve for to determine value of x at the stationary point.

(c) Take the second derivative of the the equation they gave you in part (a). Find the value of the second derivative for the value of x that you found in part (b). If this value is negative, then the stationary point is a maximum.

(d) Use the equation they gave you in part (a), with the value of x that you found in part (b). That will enable you to calculate A.

I hope that made some sense.
A question: if the second derivative of the equation is a negative constant, does that also imply that the value of x found in the previous part is the maximum?
0
9 years ago
#6
If this is in the mixed exercise then its the only question I was unable to do in the whole exercise, just the part a, can't figure it out.
0
9 years ago
#7
(Original post by Siddd)
A question: if the second derivative of the equation is a negative constant, does that also imply that the value of x found in the previous part is the maximum?
Yeah.
0
9 years ago
#8
(Original post by Siddd)
A question: if the second derivative of the equation is a negative constant, does that also imply that the value of x found in the previous part is the maximum?
Yes. The reasoning is as follows. When the first derivative is zero, the original graph must be at a stationary point. There are three possible scenarios: the stationary point could be a maximum, a minimum, or an inflection point.

Case 1: it's a maximum --
If it's a maximum, we know that the gradient of the original graph will be positive before the stationary point, and negative after the stationary point. It follows that the value of the first derivative will be positive before the stationary point, and negative after the stationary point. If you picture the graph of the first derivative based on this, you will be able to see that the gradient of the first derivative will be negative. This means that the value of the second derivative is negative. Therefore, regardless of whether or not the second derivative is constant or not: if it is negative, the original graph must be at a maximum.

Cases 2 & 3: minimum / inflection point --
These can be logically worked out, in a similar manner.

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