Thanks.
So for:
1 = p/((20-p)/2)^2 · (p-20)/2
it would be:
1 < p/((20-p)/2)^2 · (p-20)/2
... (work) ...
1 < (2p^2-40p)/((20-p)^2
p^2 - 40p + 400 < 2p^2 - 40p
0 < p^2 - 400
p > 20 or P < -20
Is that how it would look like?
Ok, never mind the post above, it was totally wrong already on the second step.
Anyways, this time I think I got it:
-1 = p/((20-p)/2)^2 · (p-20)/2
-1 = (2p^2-40p)/(20-p)^2
-(20-p)^2 = 2p^2 - 40p
-p^2 + 40p - 400 = 2p^2 - 40p
0 = 3p^2 - 80p + 400
p = 20/3, 20
Did the same...
So I tried it:
η = p / q · dq / dp
1 = p / q · dq / dp
1 = p/((20-p)/2)^2 · (p-20)/2
... (some work) ...
1 = (2p-40)/(20-p)^2
20 - p^2 = 2p - 40
0 = -60 + 2p + p^2
p = -8.81 (inadmissible)
p = 6.81
So therefore, 6.81 < p < 20.
For those values of p, η is elastic. Does it look...
Do you mean η = p / q · dq / dp
q = ((20-p)/2)^2
q' = (p-20)/2
.: η = p / ((20-p)/2)^2 · (p-20)/2
And since I know |η| must be greater than 1, then do I set η to something such as -2? or use -10? That's the only part I don't clearly understand because I could use any number, would it...
Homework Statement
Derivatives and elasticity:
The demand equation for a product is q = \left(\frac{20-p}{2}\right)^{2} for 0 \leq p \leq 20.
a) find all values of p for which demand is elastic.
Homework Equations
Elasticity: \eta = \frac{p}{q} x \frac{dq}{dp}
The Attempt at a...
Homework Statement
A real estate office manages 50 apartments in a downtown building. When the rent is $900 per month, all the units are occupied. For ever $25 increase in rent, one unit becomes vacant. On average, all units require $75 in maintenance and repairs each month. How much rent...
Homework Statement
A cylindrical shaped tin can must have a volume of 1000cm3.
Find the dimensions that require the minimum amount of tin for the can (Assume no waste material). The smallest can has a diameter of 6cm and a height of 4 cm.
Homework Equations
V = \pi r^{2}h
P = 2(...
Ok, starting from 3 again:
All I'm using is the Quotient Rule here.
=\frac{-3x^{4}}{(4x-8)^{1/2}}
=\frac{(-12x^{3})(4x-8)^{1/2} - (-3x^{4})(1/2)(4x-8)^{-1/2}(4)}{[(4x-8)^{1/2}]^{2}}
=\frac{(-12x^{3})(4x-8)^{1/2}-(-3x^{4})(1/2)(4x-8)^{-1/2}(4)}{(4x-8)}
=\frac{-12x^{3}(4x-8)^{1/2}...
Homework Statement
=\frac{-3x^{4}}{(4x-8)^{1/2}}
Is it actually correct, I'm not sure if it's correct, still.Homework Equations
Quotient Rule and Chain Rule
The Attempt at a Solution
=\frac{-3x^{4}}{(4x-8)^{1/2}}...
y = (1-x2)3 (6+2x)-3
y' = 3 (1-x2)2 (-2x)(6+2x)-3 + (1-x2)3(-3)(6+2x)-4(2)
y' = -6x (1-x2)2(6+2x)-3 - 6(1-x2)3(6+2x)-4
That's kind of the problem. I'm not sure how to take out < -6(1 - x2)2(6 + 2x)-4 >.
In the 3rd line, there is a (6+2x)-3, how do you take out (6+2x)-4? Does the power...
Thanks!
Can you help me out with one more?
y = (1-x2)3 (6+2x)-3
y' = 3 (1-x2)2 (-2x)(6+2x)-3 + (1-x2)3(-3)(6+2x)-4(2)
y' = -6x (1-x2)2(6+2x)-3 - 6(1-x2)3(6+2x)-4
Not exactly sure what to do with this.
I could possibly:
y' = -6 (1-x2)(6+2x)-3[x-(1-x2)(6+2x)-1]
or should I put...
Homework Statement
y =x4(2x-5)6
Homework Equations
Product Rule & Power of a Function Rule
The Attempt at a Solution
y = x4(2x-5)6
y' = 4(x)3(1)(2x-5)6 + x4(6)(2x-5)5(2)
y' = 4x3(2x-5)6 + 12x4(2x-5)5
The answer is:
20x3(2x-5)5(x-1)
No idea where they get the 20x3 or the...