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As Amit has commented, there is a mistake in the above expression. This however is not the answer I got as I have reworked through the math. So the answer which I think is correct along with my work is the following:

we are looking to solve $$\arg \min_{x_1,x_2} p_1x_1+p_2x_2$$ subject to: $$\bar{U}\geq-(x_1-\delta_1)^2-(x_2-\delta _2)^2$$ where $\bar{U}\leq0$.

Step 1: Set up Lagrangian $$\mathcal{L}:p_1x_1+p_2x_2+\lambda[\bar{U}+(x_1-\delta_1)^2+(x_2-\delta _2)]$$

Step 2: Take FOCs and derive optimality condition $$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_1}:p_1+2\lambda(x_1-\delta_1)=0$$ $$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_2}: p_2+2\lambda(x_2-\delta_2)=0$$

Rearranging the above we get:

$$(1a)\ \ \ \ \ \ \ p_1=-2\lambda(x_1-\delta_1)$$ $$(2a)\ \ \ \ \ \ \ p_2=-2\lambda(x_2-\delta_2)$$

Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).

$$\frac{p_2}{p_1}=\frac{x_2-\delta_2}{x_1-\delta_1}$$

Step 3: Derive "intermediate" demand equation

Rearranging our optimality condition we get:

$$\hat{x_2}=\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)$$

Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand

the following algebra below helps give our answer (apologies for the messiness).

$$\bar{U}=-(x_1-\delta_1)^2-(\hat{x_2}+\delta _2)^2$$

$$\bar{U}=-(x_1-\delta_1)^2-\left(\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)-\delta _2\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$

$$\bar{U}=-(x_1-\delta_1)^2\left(1+\left(\frac{p_2}{p_1}\right)^2\right)$$

Heres the tricky part, recall that $\bar{U}\leq 0$ this allows us to do the following:

$$(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$$ $$x_1-\delta_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$

This is allowed because the term in the brackets is a positive number.

Moving over $\delta_1$ we have our hicksian demand for $x_1$:

$$x^h_1=\delta_1+\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$ by symmetry the hicksian demand for $x_2$ is:

$$x^h_2=\delta_2+\left(\frac{-\bar{U}}{1+\left(\frac{p_1}{p_2}\right)^2}\right)^{\frac{1}{2}}$$

I hope this helps.

As Amit has commented, there is a mistake in the above expression. This however is not the answer I got as I have reworked through the math. So the answer which I think is correct along with my work is the following:

we are looking to solve $$\arg \min_{x_1,x_2} p_1x_1+p_2x_2$$ subject to: $$\bar{U}\geq-(x_1-\delta_1)^2-(x_2-\delta _2)^2$$ where $\bar{U}\leq0$.

Step 1: Set up Lagrangian $$\mathcal{L}:p_1x_1+p_2x_2+\lambda[\bar{U}+(x_1-\delta_1)^2+(x_2-\delta _2)]$$

Step 2: Take FOCs and derive optimality condition $$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_1}:p_1+2\lambda(x_1-\delta_1)=0$$ $$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_2}: p_2+2\lambda(x_2-\delta_2)=0$$

Rearranging the above we get:

$$(1a)\ \ \ \ \ \ \ p_1=-2\lambda(x_1-\delta_1)$$ $$(2a)\ \ \ \ \ \ \ p_2=-2\lambda(x_2-\delta_2)$$

Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).

$$\frac{p_2}{p_1}=\frac{x_2-\delta_2}{x_1-\delta_1}$$

Step 3: Derive "intermediate" demand equation

Rearranging our optimality condition we get:

$$\hat{x_2}=\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)$$

Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand

the following algebra below helps give our answer (apologies for the messiness).

$$\bar{U}=-(x_1-\delta_1)^2-(\hat{x_2}-\delta _2)^2$$

$$\bar{U}=-(x_1-\delta_1)^2-\left(\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)-\delta _2\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$

$$\bar{U}=-(x_1-\delta_1)^2\left(1+\left(\frac{p_2}{p_1}\right)^2\right)$$

Heres the tricky part, recall that $\bar{U}\leq 0$ this allows us to do the following:

$$(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$$ $$x_1-\delta_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$

This is allowed because the term in the brackets is a positive number.

Moving over $\delta_1$ we have our hicksian demand for $x_1$:

$$x^h_1=\delta_1+\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$ by symmetry the hicksian demand for $x_2$ is:

$$x^h_2=\delta_2+\left(\frac{-\bar{U}}{1+\left(\frac{p_1}{p_2}\right)^2}\right)^{\frac{1}{2}}$$

I hope this helps.

As Amit as commented, there is a mistake in the above expression, however- this is not the answer I got as I have reworked through the math. So the answer which I think is correct along with my work is the following:

we are looking to solve $$\arg \min_{x_1,x_2} p_1x_1+p_2x_2$$ subject to: $$\bar{U}\geq-(x_1-\delta_1)^2-(x_2-\delta _2)^2$$ where $\bar{U}\leq0$.

Step 1: Set up Lagrangian $$\mathcal{L}:p_1x_1+p_2x_2+\lambda[\bar{U}+(x_1-\delta_1)^2+(x_2-\delta _2)]$$

Step 2: Take FOCs and derive optimality condition $$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_1}:p_1+2\lambda(x_1-\delta_1)=0$$ $$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_2}: p_2+2\lambda(x_2-\delta_2)=0$$

Rearranging the above we get:

$$(1a)\ \ \ \ \ \ \ p_1=-2\lambda(x_1-\delta_1)$$ $$(2a)\ \ \ \ \ \ \ p_2=-2\lambda(x_2-\delta_2)$$

Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).

$$\frac{p_2}{p_1}=\frac{x_2-\delta_2}{x_1-\delta_1}$$

Step 3: Derive "intermediate" demand equation

Rearranging our optimality condition we get:

$$\hat{x_2}=\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)$$

Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand

the following algebra below helps give our answer (apologies for the messiness).

$$\bar{U}=-(x_1-\delta_1)^2-(\hat{x_2}+\delta _2)^2$$

$$\bar{U}=-(x_1-\delta_1)^2-\left(\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)-\delta _2\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$

$$\bar{U}=-(x_1-\delta_1)^2\left(1+\left(\frac{p_2}{p_1}\right)^2\right)$$

Heres the tricky part, recall that $\bar{U}\leq 0$ this allows us to do the following:

$$(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$$ $$x_1-\delta_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$

This is allowed because the term in the brackets is a positive number.

Moving over $\delta_1$ we have our hicksian demand for $x_1$:

$$x^h_1=\delta_1+\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$ by symmetry the hicksian demand for $x_2$ is:

$$x^h_2=\delta_2+\left(\frac{-\bar{U}}{1+\left(\frac{p_1}{p_2}\right)^2}\right)^{\frac{1}{2}}$$

I hope this helps.

As Amit has commented, there is a mistake in the above expression. This however is not the answer I got as I have reworked through the math. So the answer which I think is correct along with my work is the following:

we are looking to solve $$\arg \min_{x_1,x_2} p_1x_1+p_2x_2$$ subject to: $$\bar{U}\geq-(x_1-\delta_1)^2-(x_2-\delta _2)^2$$ where $\bar{U}\leq0$.

Step 1: Set up Lagrangian $$\mathcal{L}:p_1x_1+p_2x_2+\lambda[\bar{U}+(x_1-\delta_1)^2+(x_2-\delta _2)]$$

Step 2: Take FOCs and derive optimality condition $$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_1}:p_1+2\lambda(x_1-\delta_1)=0$$ $$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_2}: p_2+2\lambda(x_2-\delta_2)=0$$

Rearranging the above we get:

$$(1a)\ \ \ \ \ \ \ p_1=-2\lambda(x_1-\delta_1)$$ $$(2a)\ \ \ \ \ \ \ p_2=-2\lambda(x_2-\delta_2)$$

Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).

$$\frac{p_2}{p_1}=\frac{x_2-\delta_2}{x_1-\delta_1}$$

Step 3: Derive "intermediate" demand equation

Rearranging our optimality condition we get:

$$\hat{x_2}=\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)$$

Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand

the following algebra below helps give our answer (apologies for the messiness).

$$\bar{U}=-(x_1-\delta_1)^2-(\hat{x_2}+\delta _2)^2$$

$$\bar{U}=-(x_1-\delta_1)^2-\left(\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)-\delta _2\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$

$$\bar{U}=-(x_1-\delta_1)^2\left(1+\left(\frac{p_2}{p_1}\right)^2\right)$$

Heres the tricky part, recall that $\bar{U}\leq 0$ this allows us to do the following:

$$(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$$ $$x_1-\delta_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$

This is allowed because the term in the brackets is a positive number.

Moving over $\delta_1$ we have our hicksian demand for $x_1$:

$$x^h_1=\delta_1+\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$ by symmetry the hicksian demand for $x_2$ is:

$$x^h_2=\delta_2+\left(\frac{-\bar{U}}{1+\left(\frac{p_1}{p_2}\right)^2}\right)^{\frac{1}{2}}$$

I hope this helps.

As Amit as commented, there is a mistake in the above expression, however- this is not the answer I got for my work as I have reworked through it. So the answer which I think is correct along with my work is the following:

we are looking to solve $$\arg \min_{x_1,x_2} p_1x_1+p_2x_2$$ subject to: $$\bar{U}\geq-(x_1-\delta_1)^2-(x_2-\delta _2)^2$$ where $\bar{U}\leq0$.

Step 1: Set up Lagrangian $$\mathcal{L}:p_1x_1+p_2x_2+\lambda[\bar{U}+(x_1-\delta_1)^2+(x_2-\delta _2)]$$

Step 2: Take FOCs and derive optimality condition $$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_1}:p_1+2\lambda(x_1-\delta_1)=0$$ $$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_2}: p_2+2\lambda(x_2-\delta_2)=0$$

Rearranging the above we get:

$$(1a)\ \ \ \ \ \ \ p_1=-2\lambda(x_1-\delta_1)$$ $$(2a)\ \ \ \ \ \ \ p_2=-2\lambda(x_2-\delta_2)$$

Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).

$$\frac{p_2}{p_1}=\frac{x_2-\delta_2}{x_1-\delta_1}$$

Step 3: Derive "intermediate" demand equation

Rearranging our optimality condition we get:

$$\hat{x_2}=\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)$$

Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand

the following algebra below helps give our answer (apologies for the messiness).

$$\bar{U}=-(x_1-\delta_1)^2-(\hat{x_2}+\delta _2)^2$$

$$\bar{U}=-(x_1-\delta_1)^2-\left(\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)-\delta _2\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$

$$\bar{U}=-(x_1-\delta_1)^2\left(1+\left(\frac{p_2}{p_1}\right)^2\right)$$

Heres the tricky part, recall that $\bar{U}\leq 0$ this allows us to do the following:

$$(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$$ $$x_1-\delta_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$

This is allowed because the term in the brackets is a positive number.

Moving over $\delta_1$ we have our hicksian demand for $x_1$:

$$x^h_1=\delta_1+\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$ by symmetry the hicksian demand for $x_2$ is:

$$x^h_2=\delta_2+\left(\frac{-\bar{U}}{1+\left(\frac{p_1}{p_2}\right)^2}\right)^{\frac{1}{2}}$$

I hope this helps.

As Amit as commented, there is a mistake in the above expression, however- this is not the answer I got as I have reworked through the math. So the answer which I think is correct along with my work is the following:

we are looking to solve $$\arg \min_{x_1,x_2} p_1x_1+p_2x_2$$ subject to: $$\bar{U}\geq-(x_1-\delta_1)^2-(x_2-\delta _2)^2$$ where $\bar{U}\leq0$.

Step 1: Set up Lagrangian $$\mathcal{L}:p_1x_1+p_2x_2+\lambda[\bar{U}+(x_1-\delta_1)^2+(x_2-\delta _2)]$$

Step 2: Take FOCs and derive optimality condition $$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_1}:p_1+2\lambda(x_1-\delta_1)=0$$ $$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_2}: p_2+2\lambda(x_2-\delta_2)=0$$

Rearranging the above we get:

$$(1a)\ \ \ \ \ \ \ p_1=-2\lambda(x_1-\delta_1)$$ $$(2a)\ \ \ \ \ \ \ p_2=-2\lambda(x_2-\delta_2)$$

Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).

$$\frac{p_2}{p_1}=\frac{x_2-\delta_2}{x_1-\delta_1}$$

Step 3: Derive "intermediate" demand equation

Rearranging our optimality condition we get:

$$\hat{x_2}=\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)$$

Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand

the following algebra below helps give our answer (apologies for the messiness).

$$\bar{U}=-(x_1-\delta_1)^2-(\hat{x_2}+\delta _2)^2$$

$$\bar{U}=-(x_1-\delta_1)^2-\left(\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)-\delta _2\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$

$$\bar{U}=-(x_1-\delta_1)^2\left(1+\left(\frac{p_2}{p_1}\right)^2\right)$$

Heres the tricky part, recall that $\bar{U}\leq 0$ this allows us to do the following:

$$(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$$ $$x_1-\delta_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$

This is allowed because the term in the brackets is a positive number.

Moving over $\delta_1$ we have our hicksian demand for $x_1$:

$$x^h_1=\delta_1+\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$ by symmetry the hicksian demand for $x_2$ is:

$$x^h_2=\delta_2+\left(\frac{-\bar{U}}{1+\left(\frac{p_1}{p_2}\right)^2}\right)^{\frac{1}{2}}$$

I hope this helps.