1. State the problems to solve: find the total cost and average cost when unit cost is 25, solve monopoly profit with demand $y=26-2x-4x^2$ and average cost $y_c=x+8$, evaluate integrals 12a--c, find areas for problems 13 and 14, compute consumer and producer surpluses for problems 15--17, and find the profit-maximizing output and profit for problem 18. 2. Problem 1 (total and average cost from unit cost 25). Formula/rule: total cost satisfies $TC(x)=c_{unit}\cdot x$ when unit (marginal) cost is constant, and average cost $AC(x)=TC(x)/x$. Work: with unit cost 25 we have $TC(x)=25x$. Work: the average cost is $AC(x)=\dfrac{TC(x)}{x}=\dfrac{25x}{x}=25$ for $x>0$. Answer: $TC(x)=25x$ and $AC(x)=25$. 3. Problem 2 (monopolist): demand $P(x)=26-2x-4x^2$ and average cost $AC(x)=x+8$. Formula/rules: revenue $R(x)=P(x)\,x$, total cost $TC(x)=AC(x)\,x$, profit $\pi(x)=R(x)-TC(x)$, and monopolist chooses $x$ with $MR=MC$ where $MR=R'(x)$ and $MC=TC'(x)$. Work: $R(x)=x(26-2x-4x^2)=26x-2x^2-4x^3$. Work: $TC(x)=AC(x)\,x=(x+8)x=x^2+8x$. Work: $MR=R'(x)=26-4x-12x^2$ and $MC=TC'(x)=2x+8$. Work: set $MR=MC$: $26-4x-12x^2=2x+8$. Work: simplify $26-4x-12x^2-2x-8=0$ gives $18-6x-12x^2=0$ or $2x^2+x-3=0$ after dividing by -6 and rearranging. Work: solve $2x^2+x-3=0$ gives $x=1$ or $x=-3/2$, discard negative, so $x^*=1$. Work: $R(1)=26-2-4=20$, $TC(1)=1+8=9$, so $\pi(1)=20-9=11$. Answer: profit-maximizing output $x=1$ and maximum profit $11$. 4. Problem 12a. Statement: evaluate $\int_0^1(x^2-2x+3)\,dx$. Formula/rules: integrate termwise with antiderivative $\int x^2\,dx=\frac{x^3}{3}$, $\int x\,dx=\frac{x^2}{2}$. Work: $\int_0^1(x^2-2x+3)\,dx=\Big[\frac{x^3}{3}-x^2+3x\Big]_0^1=\frac{1}{3}-1+3=\frac{7}{3}$. Answer: $\dfrac{7}{3}$. 5. Problem 12b. Statement: evaluate $\int_0^1\dfrac{dx}{(2x+1)^3}$. Formula/rules: use substitution $u=2x+1$, $du=2\,dx$, so $dx=\frac{du}{2}$. Work: limits $x=0\to u=1$, $x=1\to u=3$ and integral becomes $\frac{1}{2}\int_1^3 u^{-3}\,du$. Work: $\frac{1}{2}\Big[\frac{u^{-2}}{-2}\Big]_1^3=-\frac{1}{4}\big(u^{-2}\big)_1^3=-\frac{1}{4}\big(\frac{1}{9}-1\big)=\frac{2}{9}$. Answer: $\dfrac{2}{9}$. 6. Problem 12c. Statement: evaluate $\int_a^{2a}(a^3+3ax^2+x^3)\,dx$ treating $a$ as constant. Formula/rules: integrate termwise: $\int a^3\,dx=a^3x$, $\int 3ax^2\,dx=a x^3$, $\int x^3\,dx=\frac{x^4}{4}$. Work: antiderivative $F(x)=a^3x+ax^3+\frac{x^4}{4}$. Work: $F(2a)-F(a)=\big(2a^4+8a^4+4a^4\big)-\big(a^4+a^4+\frac{a^4}{4}\big)=14a^4-\frac{9}{4}a^4=\frac{47}{4}a^4$. Answer: $\dfrac{47}{4}a^4$. 7. Problem 13. Statement: area in the first quadrant bounded by the x-axis and $y=6x+x^2-x^3$. Formula/rules: area where the curve is nonnegative for $x\ge0$ is $\int_{x_1}^{x_2} y\,dx$ between consecutive nonnegative roots. Work: solve $6x+x^2-x^3=0\Rightarrow x( -x^2+x+6)=0$ with roots $x=0,3,-2$ so first-quadrant interval is $[0,3]$. Work: area $=\int_0^3(6x+x^2-x^3)\,dx=\Big[3x^2+\frac{x^3}{3}-\frac{x^4}{4}\Big]_0^3=\frac{63}{4}$. Answer: $\dfrac{63}{4}$. 8. Problem 14. Statement: total area between parabola $y=x^2-4x$, the x-axis, and the line $x=-2$. Interpretation/rule: the finite region bounded by these is for $x\in[-2,0]$ where the parabola lies above the axis, so area is $\int_{-2}^0(x^2-4x)\,dx$. Work: $\int_{-2}^0(x^2-4x)\,dx=\Big[\frac{x^3}{3}-2x^2\Big]_{-2}^0=\frac{32}{3}$. Answer: $\dfrac{32}{3}$. 9. Problem 15 (consumer surplus). Statement: demand $p=\sqrt{9-x}$ and quantity $x_0=5$, find consumer surplus. Formula/rules: consumer surplus $CS=\int_0^{x_0} p(x)\,dx - p(x_0)\,x_0$. Work: $p(5)=\sqrt{9-5}=2$. Work: $\int_0^5\sqrt{9-x}\,dx$ via $u=9-x$ gives $\int_4^9 u^{1/2}\,du=\frac{2}{3}(9^{3/2}-4^{3/2})=\frac{2}{3}(27-8)=\frac{38}{3}$. Work: $CS=\frac{38}{3}-2\cdot5=\frac{38}{3}-10=\frac{8}{3}$. Answer: $\dfrac{8}{3}$. 10. Problem 16 (producer surplus). Statement: supply $p=(x+2)^2$ and market price $p_0=25$, find producer surplus. Formula/rules: producer surplus $PS=p_0x_0-\int_0^{x_0} supply(x)\,dx$ where $x_0$ solves $supply(x_0)=p_0$. Work: solve $(x+2)^2=25\Rightarrow x+2=\pm5$, choose nonnegative $x_0=3$. Work: $\int_0^3(x+2)^2\,dx=\Big[\frac{(x+2)^3}{3}\Big]_0^3=\frac{125-8}{3}=39$. Work: $PS=25\cdot3-39=75-39=36$. Answer: $36$. 11. Problem 17 (competitive equilibrium surpluses). Statement: demand $p=16-x^2$ and supply $p=2x+1$, find consumer and producer surplus at equilibrium. Formula/rules: equilibrium solves demand = supply; $CS=\int_0^{x_e}(d(x)-p_e)\,dx$ and $PS=\int_0^{x_e}(p_e-s(x))\,dx$ or $p_ex_e-\int_0^{x_e}s(x)\,dx$. Work: solve $16-x^2=2x+1\Rightarrow x^2+2x-15=0$ gives $x_e=3$ (positive). Work: $p_e=2\cdot3+1=7$. Work: $\int_0^3(16-x^2)\,dx=\big[16x-\frac{x^3}{3}\big]_0^3=39$ so $CS=39-7\cdot3=39-21=18$. Work: $\int_0^3(2x+1)\,dx=\big[x^2+x\big]_0^3=12$ so $PS=7\cdot3-12=21-12=9$. Answer: $CS=18$, $PS=9$. 12. Problem 18 (MR and MC, profit-maximizing output and profit). Statement: $MR=44-9x$ and assume $MC=20-7x+2x^2$ (interpreting the given expression as a cubic-capable MC), find profit-maximizing output and total profit. Formula/rules: monopolist sets $MR=MC$; revenue $R$ is antiderivative of $MR$, and $TC$ is antiderivative of $MC$, then profit $\pi=R-TC$ up to constants which cancel when measuring profit differences from zero-output baseline. Work: set $44-9x=20-7x+2x^2$ which simplifies to $2x^2+ x-12=0$. Work: solve $2x^2+x-12=0$ gives $x=3$ or $x=-4$, choose $x^*=3$. Work: integrate $MR$ to get $R(x)=\int(44-9x)\,dx=44x-\frac{9}{2}x^2$ up to constant. Work: integrate $MC$ to get $TC(x)=\int(20-7x+2x^2)\,dx=20x-\frac{7}{2}x^2+\frac{2}{3}x^3$ up to constant. Work: evaluate at $x=3$: $R(3)=44\cdot3-\frac{9}{2}\cdot9=132-\frac{81}{2}=\frac{183}{2}$. Work: $TC(3)=20\cdot3-\frac{7}{2}\cdot9+\frac{2}{3}\cdot27=60-\frac{63}{2}+18=\frac{93}{2}$. Work: profit $\pi(3)=R(3)-TC(3)=\frac{183}{2}-\frac{93}{2}=45$. Answer: profit-maximizing output $x=3$ and total profit $45$. Final answers summary: 1) $TC(x)=25x$, $AC(x)=25$. 2) Monopolist: $x^*=1$, $\pi_{max}=11$. 12a) $7/3$. 12b) $2/9$. 12c) $47/4\,a^4$. 13) $63/4$. 14) $32/3$. 15) $8/3$. 16) $36$. 17) $CS=18$, $PS=9$. 18) $x^*=3$, $\pi=45$.