Huey's Hip Pizza sells two sizes of square pizzas: a small pizza that measures $10$ inches on a side and costs $\$10$, and a large pizza that measures $15$ inches on a side and costs $\$20$. If two friends go to Huey's with $\$30$ apiece, how many more square inches of pizza can they buy if they pool their money than if they each purchase pizza alone?
My solution: I have solved that problem but however, my answer does not coincide with official answer. Here's my solution:
If they purchase alone they can buy $3$ small pizzas (in that case total area is $100\cdot 3=300$ square inches) or $1$ small+$1$ large (in that case total area is $100+225=325$ square inches). Hence when they purchase each pizza alone then the total area will be $2\cdot 300=600$ square inches or $2\cdot 325=650$.
If they purchase together they can buy:
(i) $6$ small pizzas then total area will be $6\cdot 100=600$
(ii) $1$ large + $4$ small then total area will be $225+4\cdot 100=625$
(iii) $2$ large + $2$ small then total area will be $2\cdot 225+2\cdot 100=650$
(iv) $3$ large then total area will be $3\cdot 225=675$
Eventually, the maximal difference is $675-600=75$.
Is my answer true?
Your assumption about buying $6$ small pizzas is correct but there is no incentive for that .
For $\$30$, each can get a large and a small $\quad \rightarrow 2(225+100)=2\times 325=650$
Combined, they can get $3$ large pizzas $\rightarrow 3(225)=675$
$$675-650=25$$
Your assumption about buying $6$ small pizzas is correct but there is no incentive for that .