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Uma caixa com base quadrada e sem tampa deve ter volume \(100\). Determine as
dimensões da caixa que exige a menor quantidade de material para as cinco faces.
xmhint
![[Picture]](/rendered/optimization/exercises/singVarCalcOptimization3-f80ef0d5823084e1ea734a16e90d438f.svg)
xmhintAs dimensões que resultam na ”menor
quantidade de material” para as cinco faces são as dimensões que resultam na
”menor área de superfície”. Portanto, a quantidade que precisamos ”minimizar” é a
área de superfície, \(S\).
xmhintPrecisamos ”minimizar” a área de superfície,
\(S\).
\(S=x^2 +4xy\)
Lembre-se: sem tampa!
xmhintEm seguida, expressamos \(S\) como função de, digamos, \(x\).
Para isso, precisamos expressar \(y\) em termos de \(x\).
Sabemos que \(V=100\). Portanto,
\(x^2\cdot \class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}}=100\), e
\(y=\frac {100}{\class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}}}\).
Segue que
\(S(x)=x^2 +\frac {400}{\class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}}}\).
xmhintPortanto, precisamos encontrar o mínimo (absoluto) de \(S\) em seu domínio, \((0,\class{xm-answer}{\cssId{xm-answer-4}{\phantom{\rule{7em}{1.5em}}}})\).
Precisamos encontrar os pontos críticos de \(S\). Para isso, precisamos calcular
\(S'(x)\).
xmhint\(S'(x)=2x-\frac {400}{\class{xm-answer}{\cssId{xm-answer-5}{\phantom{\rule{7em}{1.5em}}}}}\).
xmhintPrecisamos encontrar os pontos críticos de \(S\). Portanto, precisamos
resolver a equação
\(2x-\frac {400}{\class{xm-answer}{\cssId{xm-answer-6}{\phantom{\rule{7em}{1.5em}}}}}=0\).
Segue que \(S\) tem um único ponto crítico, \(x=\class{xm-answer}{\cssId{xm-answer-7}{\phantom{\rule{7em}{1.5em}}}}\).
xmhintComo
\(S'(x)=2x-\frac {400}{x^{2}}=\frac {2(x^{3}-200)}{x^{2}}\), segue que
\(S'(x)<0\) em \((0, 200^{\frac {1}{3}})\) e que \(S'(x)>0\) em \((200^{\frac {1}{3}},\infty )\).
Isso significa que \(S\) é decrescente em \((0, 200^{\frac {1}{3}})\) e crescente em \((200^{\frac {1}{3}},\infty )\). Ou seja, a função \(S\) tem tanto um
mínimo local quanto um mínimo absoluto em \(x=c\).
Portanto, as dimensões que minimizam a área de superfície são
\[ \text {largura}=\class{xm-answer}{\cssId{xm-answer-8}{\phantom{\rule{7em}{1.5em}}}},\qquad \text {comprimento}=\class{xm-answer}{\cssId{xm-answer-9}{\phantom{\rule{7em}{1.5em}}}},\qquad \text {altura}=\class{xm-answer}{\cssId{xm-answer-10}{\phantom{\rule{7em}{1.5em}}}} \]