Having trouble with this problem, can anyone help?

Question

With all other factors held constant (such as light frequency, water temperature, air temperature, nutrient supply), the photosynthetic growth, I, in grams per day, for a certain plant, hydroponically grown in artificial light, depends on the light intensity, x, measured in kilowatts, according to the formula I(x) = 180x^2 - 90x^3 . What value of x (correct to two decimal places) maximizes the photosynthetic growth?

Answer 1

You want to differentiate I(x) with respect to x once and then twice. You get:

I'(x) = 2*180*x - 3*90*x^2 = 360*x - 270*x^2

and

I''(x) = 360 - 2*270*x = 360 - 540*x

You will get maxima at x=x0 where I'(x0)=0 and I''(x0)<0. So solve,

I'(x=x0) = 360*x0 - 270*x0^2 = x0*( 360 - 270*x0) = 0

Clearly, I'(x0)=0 where x0=0 and x0=360/270=36/27=4/3. You need to evaluate the second derivative to see which one of these is actually a maxima. I''(x=0) = 360, so it cannot be a maxima because it's positive. However, I''(x=360/270) = -360, which is negative, so x=360/270=4/3 is a LOCAL maxima.

Now, NORMALLY, since x is restricted to be x>=0, you should also make sure that x=0 isn't also a local maxima that is better than x=4/3. So, you'd evaluate,

I(x=0) = 0
I(x=4/3) = 320/3

and see that x=0 isn't any better than x=4/3. However, x=0 was also an unconstrained local maxima (that is, I'(x=0)=0), so we didn't have any need to do this step. However, it's nice to know what I(x=4/3) evaluates to. Regardless, x=4/3 is what you want. So the answer is:

x = 1.33 kilowatts, giving 320/3 grams per day of photosynthetic growth.

Answer 2

maximize l(x) means first differentiation l'(x) = 0 and second derivative l"(x) to be negative
l(x) = 180x^2 - 90x^3...eqn 1
l'(x) = 360x - 270x^2 ...eqn 2
l"(x) = 360 -540x..........eqn 3
at maximum
l'(x) = 360x - 270x^2 = 90x(4-3x) = 0
either x=0 or (4-3x) = 0
x is light intensity whose nonzero value is needed
so 4-3x = 0
so 4=3x or x= 4/3
for this x to be maximum, l"(x) has to be negative at this value of x
sustitiute 4/3 for x in eqn 3
l"(x) = 360 - 540 x = 360 - 540(4/3) = 360 - 720 = - 360 which is negative
so at x = 4/3 =1.33 kw photosynthetic growth will be maximum

if growth vallue is needed, substitute x = 4/3 in eqn 1
l(x) = 180x^2 - 90x^3 = 180*[(4/3)^2] - 90*[(4/3)^3]
= 180*(16/9) - 90*(64/27)
= 320 - 213.33
= 106.67
i.e. 106.67 gms /day

Answer 3

Compute I'(x) = 360x - 270x^2.

Then get its critical numbers, where I'(x) = 0. One of them is x=0. To get the other, factor out a 90x and get

I'(x) = 90x(4 - 3x)

so the other critical number is x=4/3.

To determine whether these critical numbers correspond to maxima or minima, compute I''(x) = 360 - 540x. When I''(0) = 360, so x=0 is a local minimum. I''(4/3) = -360, so x=4/3 is a local maximum. As x -> infinity, I(x) -> -infinity, so the absolute maximum of I(x) on [0,infinity) is at the local maximum, x=4/3=1.33, to two decimal places.