Chapter 4, Problem 4.1P

a)

To determine

To find: Quantity of Twinkies and soda that maximizes utility.

Explanation of Solution

$\begin{array}{l}\text{Given utility, U = }\sqrt{\text{ts}}\\ {\text{Budget constraint, tP}}_{\text{t}}+s{P}_{\text{s}}=I\\ MRS=\frac{M{U}_t}{MUs}\\ U=\sqrt{ts}\\ M{U}_t=\frac{dU}{dt}\\ M{U}_t=\frac{1}{2}{\left(\frac{s}{t}\right)}^{1/2}\\ M{U}_s=\frac{dU}{ds}\\ M{U}_s=\frac{1}{2}{\left(\frac{s}{t}\right)}^{1/2}\end{array}$

$\begin{array}{l}\text{Budget constraint:}\\ {\text{tP}}_{\text{t}}{\text{+ sP}}_{\text{s}}\text{ = I}\\ \text{MRS = }\frac{P_t}{P_s}\\ \frac{M{U}_t}{M{U}_s}=\frac{P_t}{P_s}\\ \frac{\frac{1}{2}{\left(\frac{s}{t}\right)}^{1/2}}{\frac{1}{2}{\left(\frac{t}{s}\right)}^{1/2}}=\frac{P_t}{P_s}\\ {\left(\frac{s}{t}\right)}^{1/2}{\left(\frac{s}{t}\right)}^{1/2}=\frac{P_t}{P_s}\\ \frac{s}{t}=\frac{P_t}{P_s}\\ s=t.\frac{P_t}{P_s}\mathrm{..........}(1)\\ \text{Insert (1) budget constraint}\text{.}\\ {\text{tP}}_{\text{t}}\text{+ }\left(t.\frac{P_t}{P_s}\right){\text{P}}_{\text{s}}\text{ = I}\\ \text{t = }\frac{I}{2P_t}\\ So,\text{s = t}\text{.}\left(\frac{P_t}{P_s}\right)\\ s=\frac{I}{2P_t}.\frac{P_t}{P_s}\\ Plug{\text{ P}}_t\text{= 0}{\text{.10, P}}_s\text{= 0}\text{.25, I = 1}\\ t=\frac{I}{2P_t}=\frac{1}{2\left(0.10\right)}=5\\ s=\frac{I}{2P_s}=\frac{1}{2\left(0.25\right)}=2\end{array}$

Therefore, to maximize utility there should be 5 Twinkie and 2 soda.

b)

To determine

To find: The calculation of new allowance.

Explanation of Solution

$\begin{array}{l}\text{U = }\sqrt{\text{ts}}\\ \text{U = }\sqrt{5\times 2}\\ \text{U = }\sqrt{10}\end{array}$

After inserting the t and s function in the budget constraint:

$\begin{array}{l}U=\sqrt{\frac{I}{2P_t}.\frac{I}{2P_s}}\\ U=\frac{I}{2\sqrt{P_t.P_s}}\\ \text{Plug U = }\sqrt{10},{\text{P}}_t=0.40,{\text{P}}_s=0.25\\ U=\frac{I}{2\sqrt{P_t.P_s}}\\ \sqrt{10}=\frac{I}{2\sqrt{\left(0.40\right)\left(0.25\right)}}\\ \text{Solve for the I,}\\ \text{I = 2}\end{array}$

The allowance should increase from $ 1 to $2.