MHB Calculate Change in Q(K,L) w/ Partial Derivatives Given

[Orion2]

Production function Q(K,L) without equation

However partial derivatives are given

Partial derivatives:

Q(K,L) = (K^2 - KL + L^2)/(K+L) + 4K . ln(K+L) Derivative to K
Q(K,L) =( K^2 + L^2) / (K+ L) Dervative to L

A. Calculate the derivative in point (10,L)

If I am correct this is Q'(10,L) = (100 + L^2) / (10 + L)

B. Calculate the change in Produced units when K remains 10 and L changes from 5 tot 10.

I know: Delta Q = Integral (from 5 to 10) [(100 + L^2) / (L + 10) dL]

This will give the right answer. But we have to calculate it with a simple calculator and I can't work out this integral by hand. Is there a alternative way to calculate this?

Thanks!

 

[I like Serena]

Production function Q(K,L) without equation

However partial derivatives are given

Partial derivatives:

Q(K,L) = (K^2 - KL + L^2)/(K+L) + 4K . ln(K+L) Derivative to K
Q(K,L) =( K^2 + L^2) / (K+ L) Dervative to L

A. Calculate the derivative in point (10,L)

If I am correct this is Q'(10,L) = (100 + L^2) / (10 + L)

Hi Orion! Welcome to MHB! ;)

The derivative would be both partial derivatives.
That is:
$$DQ(10, L) = \begin{pmatrix}(100 - 10L + L^2)/(10+L) + 40 \ln(10+L) \\ ( 100 + L^2) / (10+ L) \end{pmatrix}$$

B. Calculate the change in Produced units when K remains 10 and L changes from 5 tot 10.

I know: Delta Q = Integral (from 5 to 10) [(100 + L^2) / (L + 10) dL]

This will give the right answer. But we have to calculate it with a simple calculator and I can't work out this integral by hand. Is there a alternative way to calculate this?

Thanks!

We can approximate it based on the partial derivative:
$$\Delta Q = \int_5^{10} \frac{100 + L^2}{L + 10} \, dL \approx \frac{100 + L^2}{L + 10} \cdot (10-5) $$

Or we can calculate it exactly by substituting $u=L+10 \Rightarrow L=u-10 \Rightarrow dL=du$:
$$\Delta Q = \int_5^{10} \frac{100 + L^2}{L + 10} \, dL
= \int_{15}^{20} \frac{100 + (u-10)^2}{u}\,du
= \int_{15}^{20} \left(\frac{200}{u} + u - 20\right)\,du
= \left.200 \ln u + \frac 12u^2 - 20u \right|_{15}^{20}
$$