Question (Applications of Differentiation)

a. Now, consider the cardboard: xy = 3600

y = 3600 / x

Consider the picture:

(y - 12 - 13)(x - 8 - 8) = A

(y - 25)(x - 16) = A

xy - 25x - 16y + 400 = A

3600 - 25x - 16(3600 / x) + 400 = A

A = -25x + 4000 - 57 600/x


b. dA/dx = -25 + 57 600/x2

d2A/dx2 = -115 200/x3

Set dA/dx = 0

-25 + 57 600/x2 = 0

x = 48 or -48 (rejected) (Since x > 0)

When x = 48, d2A/dx2 < 0

So, max. of A is obtained when x = 48

So, the larger value of A

= -25(48) + 4000 - 57 600/48

= 1600


c.i. As A attains a maximum when x = 48.

And there is only one turning point.

So, for x > 48, A decreases as x increases.


ii. For x >= 50, A decreases as x increases.

So, A is largest when x = 50

that is, max. of A

= -25(50) + 4000 - 57 600/50

= 1598


d. 4/9 < x/y < 9/16

4y/9 < x < 9y/16

4/9 (3600/x) < x < 9/16 (3600/x)

1600/x < x < 2025/x

1600 < x2 < 2025

40 < x < 45 (Since x > 0)


For this region, A increases as x increases.

So, the largest value of A is obtained when x = 45

Largest value of A

= -25(45) + 4000 - 57 600/45

= 1595

a)
A = (x - 16)(y -25). But xy = 3600, thus
A = (x - 16)(3600/x - 25) = 3600 - 25x - 57600/x + 400
= 4000 - 25x - 57600/x.
b)
dA/dx = -25 + 57600/x^2. Put it to zero, we get
x^2 = 57600/25 = (240/5)^2
x = 240/5 = 48, which is a max. since d^2A/dx^2 < 0.
So A max. = (48 - 16)(3600/48 - 25) = 32 x 50 = 1600.
c) When A = (x -16)(y -25) = 0.
x = 16 or y = 25 (that is x = 3600/25 = 144).
So for A decreases as x increases, 48 < x < 144.
d)
4/9 < x/y < 9/16 and xy = 3600, that is
4/9 < x^2/3600 < 9/16
1600< x^2 < 2025
40< x < 45 since x is positive.
Largest value of A is when x = 45 because A increase as x increases. That is A = (45 - 16)(80 - 25) = 29 x 55 =1595.