Answer:
V = 23.85 ft³
Explanation:
The dimension of the square base = x ft by x ft
Height of box = y
Volume of the box,
![V = x^(2) y](https://img.qammunity.org/2021/formulas/mathematics/college/mc7phqwwrzjiimjko3f7fhzpvfslrrq885.png)
Cost of top = $4 *
![x^(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/8h1y97mwj8mwic83188qjkhvavc5d1kao6.png)
Cost of the bottom = $9 *
![x^(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/8h1y97mwj8mwic83188qjkhvavc5d1kao6.png)
Cost of the sides = $3 * 4xy
Total cost = 4x² + 9x² + 12xy
Total cost = 13x² + 12xy
Total cost = $193
193 = 13x² + 12xy
![y = (193 - 13x^(2) )/(12x)](https://img.qammunity.org/2021/formulas/mathematics/college/r9og26wx6mbp7fksj1ekfa7fwwmj4w9vik.png)
But volume, V = x²y
V =
![(x^(2) (193 - 13x^(2)) )/(12x)](https://img.qammunity.org/2021/formulas/mathematics/college/zcut2gg9nwkxn59ebjnes7bl7nv4q72aji.png)
V =
...................(1)
At maximum value, V' = 0
0 =
![(193 - 39x^(2) )/(12)](https://img.qammunity.org/2021/formulas/mathematics/college/lkubuocpqo2yqml40d0pt5ed6viodnresp.png)
![39x^(2) = 193\\x^(2) = (193)/(39)](https://img.qammunity.org/2021/formulas/mathematics/college/h8nn0fqq6tar6q6qzruw3wsylo9672pczt.png)
x = 2.23
Put the value of x into (1)
![V = (2.23 (193 - 13*2.23^(2)) )/(12)](https://img.qammunity.org/2021/formulas/mathematics/college/dk2q6sg6tf2ufvvfiz512uoth8e0mtiks1.png)
V = 23.85 ft³